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These Are the 7 Hardest Math Problems Ever Solved — Good Luck in Advance

On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?

That turned out to be much harder—as in, no one was able to solve for those integers for 65 years until a supercomputer finally came up with the solution to 42. (For the record: x = -80538738812075974, y = 80435758145817515, and z = 12602123297335631. Obviously.)

That’s the beauty of math: There’s always an answer for everything, even if takes years, decades, or even centuries to find it. So here are seven more brutally difficult math problems that once seemed impossible until mathematicians found a breakthrough.

Henri Poincaré was a French mathematician who, around the turn of the 20th century, did foundational work in what we now call topology. Here’s the idea: Topologists want mathematical tools for distinguishing abstract shapes. For shapes in 3D space, like a ball or a donut, it wasn’t very hard to classify them all . In some significant sense, a ball is the simplest of these shapes.

Poincaré then went up to 4-dimensional stuff, and asked an equivalent question. After some revisions and developments, the conjecture took the form of “Every simply-connected, closed 3-manifold is homeomorphic to S^3,” which essentially says “the simplest 4D shape is the 4D equivalent of a sphere.”

Still with us?

A century later, in 2003, a Russian mathematician named Grigori Perelman posted a proof of Poincaré’s conjecture on the modern open math forum arXiv. Perelman’s proof had some small gaps, and drew directly from research by American mathematician Richard Hamilton. It was groundbreaking, yet modest.

After the math world spent a few years verifying the details of Perelman’s work, the awards began . Perelman was offered the million-dollar Millennium Prize, as well as the Fields Medal, often called the Nobel Prize of Math. Perelman rejected both. He said his work was for the benefit of mathematics, not personal gain, and also that Hamilton, who laid the foundations for his proof, was at least as deserving of the prizes.

It’s a simple one to write. There are many trios of integers (x,y,z) that satisfy x²+y²=z². These are known as the Pythagorean Triples, like (3,4,5) and (5,12,13). Now, do any trios (x,y,z) satisfy x³+y³=z³? The answer is no, and that’s Fermat’s Last Theorem.

Fermat famously wrote the Last Theorem by hand in the margin of a textbook, along with the comment that he had a proof, but could not fit it in the margin. For centuries, the math world has been left wondering if Fermat really had a valid proof in mind.

Flash forward 330 years after Fermat’s death to 1995, when British mathematician Sir Andrew Wiles finally cracked one of history’s oldest open problems . For his efforts, Wiles was knighted by Queen Elizabeth II and was awarded a unique honorary plaque in lieu of the Fields Medal, since he was just above the official age cutoff to receive a Fields Medal.

Wiles managed to combine new research in very different branches of math in order to solve Fermat’s classic number theory question. One of these topics, Elliptic Curves, was completely undiscovered in Fermat’s time, leading many to believe Fermat never really had a proof of his Last Theorem.

Grab any map and four crayons. It’s possible to color each state (or country) on the map, following one rule: No states that share a border get the same color.

The fact that any map can be colored with five colors—the Five Color Theorem —was proven in the 19th century. But getting that down to four took until 1976.

Two mathematicians at the University of Illinois, Urbana-Champaign, Kenneth Appel and Wolfgang Hakan, found a way to reduce the proof to a large, finite number of cases . With computer assistance, they exhaustively checked the nearly 2,000 cases, and ended up with an unprecedented style of proof.

Arguably controversial since it was partially conceived in the mind of a machine, Appel and Hakan’s proof was eventually accepted by most mathematicians. It has since become far more common for proofs to have computer-verified parts, but Appel and Hakan blazed the trail.

Cantor proved that the set of real numbers is larger than the set of natural numbers, which we write as |ℝ|>|ℕ|. It was easy to establish that the size of the natural numbers, |ℕ|, is the first infinite size; no infinite set is smaller than ℕ.

Now, the real numbers are larger, but are they the second infinite size? This turned out to be a much harder question, known as The Continuum Hypothesis (CH) .

If CH is true, then |ℝ| is the second infinite size, and no infinite sets are smaller than ℝ, yet larger than ℕ. And if CH is false, then there is at least one size in between.

So what’s the answer? This is where things take a turn.

CH has been proven independent, relative to the baseline axioms of math. It can be true, and no logical contradictions follow, but it can also be false, and no logical contradictions will follow.

It’s a weird state of affairs, but not completely uncommon in modern math. You may have heard of the Axiom of Choice, another independent statement. The proof of this outcome spanned decades and, naturally, split into two major parts: the proof that CH is consistent, and the proof that the negation of CH is consistent.

The first half is thanks to Kurt Gödel, the legendary Austro-Hungarian logician. His 1938 mathematical construction, known as Gödel’s Constructible Universe , proved CH compatible with the baseline axioms, and is still a cornerstone of Set Theory classes. The second half was pursued for two more decades until Paul Cohen, a mathematician at Stanford, solved it by inventing an entire method of proof in Model Theory known as “forcing.”

Gödel’s and Cohen’s halves of the proof each take a graduate level of Set Theory to approach, so it’s no wonder this unique story has been esoteric outside mathematical circles.

The Prime Number Theorem is more subtle; it describes the distribution of prime numbers along the number line. More precisely, it says that, given a natural number N, the number of primes below N is approximately N/log(N) ... with the usual statistical subtleties to the word “approximately” there.

Drawing on mid-19th-century ideas, two mathematicians, Jacques Hadamard and Charles Jean de la Vallée Poussin, independently proved the Prime Number Theorem in 1898. Since then, the proof has been a popular target for rewrites, enjoying many cosmetic revisions and simplifications. But the impact of the theorem has only grown.

The usefulness of the Prime Number Theorem is huge. Modern computer programs that deal with prime numbers rely on it. It’s fundamental to primality testing methods, and all the cryptology that goes with that.

Now, if we go up to ax³+bx²+cx+d=0, a closed form for “x=” is possible to find, although it’s much bulkier than the quadratic version. It’s also possible, yet ugly, to do this for degree 4 polynomials ax⁴+bx³+cx²+dx+f=0.

The goal of doing this for polynomials of any degree was noted as early as the 15th century. But from degree 5 on, a closed form is not possible. Writing the forms when they’re possible is one thing, but how did mathematicians prove it’s not possible from 5 up?

The world was only starting to comprehend the brilliance of French mathematician Evariste Galois when he died at the age of 20 in 1832. His life included months spent in prison, where he was punished for his political activism, writing ingenious, yet unrefined mathematics to scholars, and it ended in a fatal duel.

Galois’ ideas took decades after his death to be fully understood, but eventually they developed into an entire theory now called Galois Theory . A major theorem in this theory gives exact conditions for when a polynomial can be “solved by radicals,” meaning it has a closed form like the quadratic formula. All polynomials up to degree 4 satisfy these conditions, but starting at degree 5, some don’t, and so there’s no general form for a solution for any degree higher than 4.

The Ancient Greeks wondered about constructing lines and shapes in various ratios, using the tools of an unmarked compass and straightedge. If someone draws an angle on some paper in front of you, and gives you an unmarked ruler, a basic compass, and a pen, it’s possible for you to draw the line that cuts that angle exactly in half. It’s a quick four steps, nicely illustrated like this , and the Greeks knew it two millennia ago.

What eluded them was cutting an angle in thirds. It stayed elusive for literally 15 centuries, with hundreds of attempts in vain to find a construction. It turns out such a construction is impossible.

Modern math students learn the angle trisection problem—and how to prove it’s not possible—in their Galois Theory classes. But, given the aforementioned period of time it took the math world to process Galois’ work, the first proof of the problem was due to another French mathematician, Pierre Wantzel . He published his work in 1837, 16 years after the death of Galois, but nine years before most of Galois’ work was published.

Either way, their insights are similar, casting the construction question into one about properties of certain representative polynomials. Many other ancient construction questions became approachable with these methods, closing off some of the oldest open math questions in history.

So if you ever time-travel to ancient Greece, you can tell them their attempts at the angle trisection problem are futile.

In 2019, mathematicians finally solved a math puzzle that had stumped them for decades. It’s called a Diophantine Equation, and it’s sometimes known as the “summing of three cubes”: Find x, y, and z such that x³+y³+z³=k, for each k from one to 100.

These ten brutally difficult math problems once seemed impossible until mathematicians eventually solved them—even if it took them years, decades, or centuries.

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10 World’s Hardest Math Problems With Solutions and Examples That Will Blow Your Mind

  • Updated: August 31, 2023
  • Category: Advanced Math , Grade 3 , Grade 4 , Grade 5 , Grade 6

WORLD’S HARDEST MATH PROBLEMS

Update : This article was last updated on 12th Oct 2023 to reflect the accuracy and up-to-date information on the page.

The mystical world of mathematics—is home to confounding problems that can make even the most seasoned mathematicians scratch their heads. Yet, it’s also a realm where curiosity and intellect shine the brightest. 

Here are 10 of the world’s hardest math problems , with solutions and examples for those that are solved and a humble “unsolved” tag for the puzzles that continue to confound experts.

1. The Four Color Theorem

The Four Color Theorem

Source : Research Outreach

Problem : Can every map be colored with just four colors so that no two adjacent regions have the same color?

Status : Solved

Solution Example : The Four Color Theorem was proven with computer assistance, checking numerous configurations to show that four colors are sufficient. If you want to prove it practically, try coloring a map using only four colors; you’ll find it’s always possible without adjacent regions sharing the same color.

2. Fermat’s Last Theorem

Problem : There are no three positive integers a,b,c that satisfies

a n +b n =c n for n>2.

Solution Example : Andrew Wiles provided a proof in 1994. To understand it, one would need a deep understanding of elliptic curves and modular forms. The proof shows that no such integers a,b,c can exist for n>2.

3. The Monty Hall Problem

The Monty Hall Problem

Source:  Towards Data Science

Problem : You’re on a game show with three doors. One hides a car, the others goats. After choosing a door, the host reveals a goat behind another door. Do you switch?

Solution Example: Always switch. When you first choose, there’s a 1/3 chance of picking the car. After a goat is revealed, switching gives you a 2/3 chance of winning. If you don’t believe it, try simulating the game multiple times.

4. The Travelling Salesman Problem

The Travelling Salesman Problem

Source : Brilliant

Problem: What’s the shortest possible route that visits each city exactly once and returns to the origin?

Status: Unsolved for a general algorithm

Solution Example: This is known as computer science’s most well-known optimization problems. Although there is no solution for all cases, algorithms like the Nearest Neighbor and Dynamic Programming can provide good approximations for specific instances.

5. The Twin Prime Conjecture

The Twin Prime Conjecture

Source: Hugin

Problem : Are there infinitely many prime numbers that differ by 2?

Status : Unsolved

Solution Example : N/A

Recommended Reading: Pros and Cons of Math Competition

6. The Poincaré Conjecture

The Poincaré Conjecture

Problem : Can every simply connected, closed 3-manifold be homomorphic to the 3-sphere?

Solution Example : Grigori Perelman proved this using Richard Hamilton’s Ricci flow program. In simple terms, he showed that every shape meeting the problem’s criteria can be stretched and shaped into a 3-sphere.

7. The Goldbach Conjecture

The Goldbach Conjecture

Source: Medium

Problem : Can every even integer greater than 2 be expressed as the sum of two prime numbers ?

8. The Riemann Hypothesis

The Riemann Hypothesis

Source:  The Aperiodical

Problem : Do all non-trivial zeros of the Riemann zeta function have their real parts equal to 1/2?

9. The Collatz Conjecture

The Collatz Conjecture

Source: Python in Plain English

Problem : Starting with any positive integer n, the sequence n,n/2,3n+1,… eventually reaches 1.

Status: Unsolved

Solution Example: N/A

10. Navier–Stokes Existence and Smoothness

Navier–Stokes Existence and Smoothness

Problem: Do solutions to the Navier–Stokes equations exist, and are they smooth?

There you have it—10 of the world’s hardest math problems. Some have been gloriously solved, giving us a brilliant glimpse into the capabilities of human intellect. Others still taunt the academic world with their complexity. For math lovers, this is the playground that never gets old, the arena where they can continually hone their problem-solving skills. So, do you feel up to the challenge?

Moonpreneur understands the needs and demands this rapidly changing technological world is bringing with it for our kids. Our expert-designed Advanced Math course for grades 3rd, 4th, 5th, and 6th will help your child develop math skills with hands-on lessons, excite them to learn, and help them build real-life applications. 

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does solving hard math questions even sharpen your brain? I mean I get exhausted and start panicking

Simran Chawla

Yes, solving tough math problems sharpens your brain by boosting critical thinking and problem-solving skills. While it may feel challenging, the process enhances cognitive abilities over time.

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Don’t panic, take breaks, have some time, you’ll get better eventually, not immediately.

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Do you know there is an award price of $1 Million for solving the Riemann Hypothesis….. THIS IS SOMETHING SOO SERIOUS!!!!

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13 World’s Hardest Math Problems | With Solutions

worlds hardest math problem solved

For decades, mathematics has been a fascinating and challenging topic. People have been interested in learning and getting good at math from ancient Greeks to modern mathematicians. But have you ever wondered which math problem is the most challenging?

What could be so tricky and complicated that only some of the brightest mathematicians have been able to solve it? This article will look at 13 of the hardest math problems and how mathematicians have tried to solve them.

Continue reading the article to explore the world’s hardest math problems, listed below.

The Poincaré Conjecture

The prime number theorem, fermat’s last theorem, the reimann hypothesis, classification of finite simple groups, four color theorem, goldbach’s conjecture.

  • Inscribed Square Problem

Twin Prime Conjecture

The continuum hypothesis, collatz conjecture, birch and swinnerton-dyer conjecture, the kissing number problem.

worlds hardest math problem solved

Mathematicians struggled for about a century with the Poincaré conjecture, which was put forth by Henri Poincaré in 1904.

According to this theory,

every closed, connected three-dimensional space is topologically identical to a three-dimensional sphere (S3).

We must explore the field of topology to comprehend what this entails. The study of properties of objects that hold after being stretched, bent, or otherwise distorted is known as topology. In other words, topologists are fascinated by how things can change without rupturing or being torn.

The topology of three-dimensional spaces is the subject of the Poincaré conjecture. A space volume with three dimensions—length, breadth, and height—is a three-dimensional space. A three-dimensional object called a sphere has a round and curved surface.

According to the Poincaré Conjecture, a three-sphere (S3), or the collection of points in four dimensions that are all at a fixed distance from a given point, is topologically identical to every simply-connected, closed, three-dimensional space (i.e., one that has no gaps or voids) and edges. 

Although it would appear easy, it took more than a century to confirm the conjecture thoroughly.

  • Poincaré expanded his hypothesis to include any dimension (n-sphere). 
  • Stephen Smale, an American mathematician, proved the conjecture to be true for n = 5 in 1961.
  • Freedman, another American mathematician, proved the conjecture to be true for n = 4 in 1983. 
  • Grigori Perelman, a Russian mathematician, then proved the conjecture to be true for n = 3 in 2002, completing the solution.
  • Perelman eventually addressed the problem by combining topology and geometry. One of the highest awards in mathematics, the Fields Medal, was given to all three mathematicians. Perelman rejected the Fields Medal. He was also given a $1 million prize by the Clay Mathematics Institute (CMI) of Cambridge, Massachusetts, for resolving one of the seven Millennium Problems, considered one of the world’s most challenging mathematical puzzles. However, he turned it down as well.

The prime number theorem (PNT) explains how prime numbers asymptotically distribute among positive integers. It shows how fast primes become less common as numbers get bigger.

The prime number theorem states that the number of primes below a given natural number N is roughly N/log(N), with the word “approximately” carrying the typical statistical connotations.

  • Two mathematicians, Jacques Hadamard and Charles Jean de la Vallée Poussin, independently proved the Prime Number Theorem in 1896. Since then, the proof has frequently been the subject of rewrites, receiving numerous updates and simplifications. However, the theorem’s influence has only increased.

French lawyer and mathematician Pierre de Fermat lived in the 17th century. Fermat was one of the best mathematicians in history. He talked about many of his theorems in everyday conversation because math was more of a hobby for him.

He made claims without proof, leaving it to other mathematicians decades or even centuries later to prove them. The hardest of them is now referred to as Fermat’s Last Theorem.

Fermat’s last theorem states that;

there are no positive integers a, b, and c that satisfy the equation an + bn = cn for any integer value of n greater than 2.
  • In 1993, British mathematician Sir Andrew Wiles solved one of history’s longest mysteries. As a result of his efforts, Wiles was knighted by Queen Elizabeth II and given a special honorary plaque rather than the Fields Medal because he was old enough to qualify.
  • Wiles synthesized recent findings from many distinct mathematics disciplines to find answers to Fermat’s well-known number theory query.
  • Many people think Fermat never had proof of his Last Theorem because Elliptic Curves were utterly unknown in Fermat’s time.

worlds hardest math problem solved

Mathematicians have been baffled by the Riemann Hypothesis for more than 150 years. It was put forth by the German mathematician Bernhard Riemann in 1859. According to Riemann’s Hypothesis

Every Riemann zeta function nontrivial zero has a real component of ½.

The distribution of prime numbers can be described using the Riemann zeta function. Prime numbers, such as 2, 3, 5, 7, and 11, can only be divided by themselves and by one. Mathematicians have long been fascinated by the distribution of prime numbers because figuring out their patterns and relationships can provide fresh perspectives on number theory and other subject areas.

Riemann’s hypothesis says there is a link between how prime numbers are spread out and how the zeros of the Riemann zeta function are set up. If this relationship is accurate, it could significantly impact number theory and help us understand other parts of mathematics in new ways.

  • The Riemann Hypothesis is still unproven, despite being one of mathematics’ most significant unsolved issues.
  • Michael Atiyah, a mathematician, proclaimed in 2002 that he had proved the Riemann Hypothesis, although the mathematical community still needs to acknowledge his claim formally.
  • The Clay Institute has assigned the hypothesis as one of the seven Millennium Prize Problems. A $1 million prize is up for anyone who can prove the Riemann hypothesis to be true or false.

Abstract algebra can be used to do many different things, like solve the Rubik’s cube or show a body-swapping fact in Futurama. Algebraic groups follow a few basic rules, like having an “identity element” that adds up to 0. Groups can be infinite or finite, and depending on your choice of n, it can be challenging to describe what a group of a particular size n looks like.

There is one possible way that the group can look at whether n is 2 or 3. There are two possibilities when n equals 4. Mathematicians intuitively wanted a complete list of all feasible groups for each given size.

  • The categorization of finite simple groups, arguably the most significant mathematical undertaking of the 20th century, was planned by Harvard mathematician Daniel Gorenstein, who presented the incredibly intricate scheme in 1972.
  • By 1985, the project was almost finished, but it had consumed so many pages and publications that peer review by a single person was impossible. The proof’s numerous components were eventually reviewed one by one, and the classification’s completeness was verified.
  • The proof was acknowledged mainly by the 1990s. Verification was later streamlined to make it more manageable, and that project is still active today.

worlds hardest math problem solved

According to four color theorem

Any map in a plane can be given a four-color coloring utilizing the rule that no two regions sharing a border (aside from a single point) should have the same color.
  • Two mathematicians at the University of Illinois at Urbana-Champaign, Kenneth Appel and Wolfgang Hakan identified a vast, finite number of examples to simplify the proof. They thoroughly examined the over 2,000 cases with the aid of computers, arriving at an unheard-of proof style.
  • The proof by Appel and Hakan was initially debatable because a computer generated it, but most mathematicians ultimately accepted it. Since then, there has been a noticeable increase in the usage of computer-verified components in proofs, as Appel and Hakan set the standard.

worlds hardest math problem solved

According to Goldbach’s conjecture, every even number (higher than two) is the sum of two primes. You mentally double-check the following for small numbers: 18 is 13 + 5, and 42 is 23 + 19. Computers have tested the conjecture for numbers up to a certain magnitude. But for all natural numbers, we need proof.

Goldbach’s conjecture resulted from correspondence between Swiss mathematician Leonhard Euler and German mathematician Christian Goldbach in 1742.

  • Euler is regarded as one of the finest mathematicians in history. Although I cannot prove it, in the words of Euler, “I regard [it] as a totally certain theorem.”
  • Euler might have understood why it is conversely tricky to resolve this problem. More significant numbers have more methods than smaller ones to be expressed as sums of primes. In the same way that only 3+5 can split eight into two prime numbers, 42 can be divided into 5+37, 11+31, 13+29, and 19+23. Therefore, for vast numbers, Goldbach’s Conjecture is an understatement.
  • The Goldbach conjecture has been confirmed for all integers up to 4*1018, but an analytical proof has yet to be found.
  • Many talented mathematicians have attempted to prove it but have yet to succeed.

Inscribed Sq uare Problem

Another complex geometric puzzle is the “square peg problem,” also known as the “inscribed square problem” or the “Toeplitz conjecture.” The Inscribe Square Problem Hypothesis asks:

Does every simple closed curve have an inscribed square?

In other words, it states, ” For any curve, you could draw on a flat page whose ends meet (closed), but lines never cross (simple); we can fit a square whose four corners touch the curve somewhere.

  • The inscribed square problem is unsolved in geometry.
  • It bears the names of mathematicians Bryan John Birch and Peter Swinnerton-Dyer, who established the conjecture using automated calculation in the first half of the 1960s.
  • Only specific instances of the hypothesis have been proven as of 2023.

The Twin Prime Conjecture is one of many prime number-related number theory puzzles. Twin primes are two primes that differ from each other by two. The twin prime examples include 11 and 13 and 599 and 601. Given that there are an unlimited number of prime numbers, according to number theory, there should also be an endless number of twin primes.

The Twin Prime Conjecture asserts that there are limitless numbers of twin primes.

  • In 2013, Yitang Zhang did groundbreaking work to solve the twin prime conjecture.
  • However, the twin prime conjecture still needs to be solved.

Infinities are everywhere across modern mathematics. There are infinite positive whole numbers (1, 2, 3, 4, etc.) and infinite lines, triangles, spheres, cubes, polygons, etc. It has also been proven by modern mathematics that there are many sizes of infinity.

If the elements of a set can be arranged in a 1-to-1 correspondence with the positive whole numbers, we say the set of elements is countably infinite. Therefore, the set of whole numbers and rational numbers are countable infinities.

Georg Cantor found that the set of real numbers is uncountable. In other words, even if we used all the whole numbers, we would never be able to go through and provide a positive whole number to every real number. Uncountable infinities might be seen as “larger” than countable infinities.

  • According to the continuum hypothesis, there must be a set of numbers whose magnitude strictly falls between countably infinite and uncountably infinite. The continuum hypothesis differs from the other problems in this list in that it is impossible to solve or at least impossible to address using present mathematical methods.
  • As a result, even though we have yet to determine whether the continuum hypothesis is accurate, we do know that it cannot be supported by the tools of modern set theory either. It would be necessary to develop a new framework for set theory, which has yet to be done, to resolve the continuum hypothesis.

worlds hardest math problem solved

To understand Collatz’s conjecture, try to understand the following example. First, you have to pick a positive number, n. Then, from the last number, create the following sequence:

If the number is even, divide by 2. If it’s odd, multiply by 3 and then add 1. The objective is to keep going through this sequence until you reach 1. Let’s try this sequence with the number 12 as an example. Starting with number 12, we get: 12, 6, 3, 10, 5, 16, 8, 4, 2, 1

Starting at 19, we obtain the following: 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1

According to the Collatz conjecture, this sequence will always end in 1, regardless of the value of n you started with. This speculation has been tested for all values of n up to 87,260, but no proof has been found.

  • Collatz’s conjecture has been unsolved up till now.
  • Mathematical problem-solver Paul Erdree once said of the Collatz Conjecture, “Mathematics may not be ready for such problems.”

Two British mathematicians, Bryan Birch and Peter Swinnerton-Dyer formulated their hypotheses in the 1960s. The Birch and Swinnerton-Dyer conjecture in mathematics describes rational answers to the equations defining an elliptic curve.

This hypothesis states explicitly that there are an infinite number of rational points (solutions) if ζ(1) equals 0 and that there are only a finite number of such places if ζ(1) is not equal to 0.

  • For Birch and Swinnerton-Dyer’s conjecture, Euclid provided a comprehensive solution, but this becomes very challenging for problems with more complex solutions.
  • Yu. V. Matiyasevich demonstrated in 1970 that Hilbert’s tenth problem could not be solved, saying there is no mechanism for identifying when such equations have a whole number solution.
  • As of 2023, only a few cases have been solved.

worlds hardest math problem solved

Each sphere has a Kissing Number, the number of other spheres it is kissing, when a group of spheres is packed together in one area. For example, your kissing number is six if you touch six nearby spheres. Nothing difficult.

Mathematically, the condition can be described by the average kissing number of a tightly packed group of spheres. However, a fundamental query regarding the kissing number remains unsolved.

First, you must learn about dimensions to understand the kissing number problem. In mathematics, dimensions have a special meaning as independent coordinate axes. The two dimensions of a coordinate plane are represented by the x- and y-axes. 

A line is a two-dimensional object, whereas a plane is a three-dimensional object. Mathematicians have established the highest possible kissing number for spheres with those few dimensions for these low numbers. On a 1-D line, there are two spheres—one to your left and the other to your right.

  • The Kissing Problem is generally unsolved in dimensions beyond three.
  • A complete solution for the kissing problem number faces many obstacles, including computational constraints. The debate continued to solve this problem.

The Bottom Line

When it comes to pushing the boundaries of the enormous human ability to comprehend and problem-solving skills, the world’s hardest math problems are unquestionably the best. These issues, which range from the evasive Continuum Hypothesis to the perplexing Riemann Hypothesis, continue to puzzle even the sharpest mathematicians.

But regardless of how challenging they are, these problems keep mathematicians inspired and driven to explore new frontiers. Whether or not these problems ever get resolved, they illustrate the enormous ability of the human intellect.

Even though some of these issues might never fully be resolved, they continue to motivate and inspire advancement within the field of mathematics and reflects how broad and enigmatic this subject is!

Let us know out of these 13 problems which problem you find the hardest!

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5 of the world’s toughest unsolved maths problems

The Open Problems in Mathematical Physics is a list of the most monstrous maths riddles in physics. Here are five of the top problems that remain unsolved

By Benjamin Skuse

7 February 2019

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1. Separatrix Separation

A pendulum in motion can either swing from side to side or turn in a continuous circle. The point at which it goes from one type of motion to the other is called the separatrix, and this can be calculated in most simple situations. When the pendulum is prodded at an almost constant rate though, the mathematics falls apart. Is there an equation that can describe that kind of separatrix?

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2. Navier–Stokes

The Navier-Stokes equations, developed in 1822, are used to describe the motion of viscous fluid. Things like air passing over an aircraft wing or water flowing out of a tap. But there are certain situations in which it is unclear whether the equations fail or give no answer at all. Many mathematicians have tried – and failed – to resolve the matter, including Mukhtarbay Otelbaev of the Eurasian National University in Astana, Kazakhstan. In 2014, he claimed a solution, but later retracted it. This is one problem that is worth more than just prestige. It is also one of the Millennium Prize Problems , which means anyone who solves it can claim $1 million in prize money.

Read more: The baffling quantum maths solution it took 10 years to understand

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3. Exponents and dimensions

Imagine a squirt of perfume diffusing across a room. The movement of each molecule is random, a process called Brownian motion, even if the way the gas wafts overall is predictable. There is a mathematical language that can describe things like this, but not perfectly. It can provide exact solutions by bending its own rules or it can remain strict, but never quite arrive at the exact solution. Could it ever tick both boxes? That is what the exponents and dimensions problem asks. Apart from the quantum Hall conductance problem , this is the only one on the list that is at least partially solved. In 2000, Gregory Lawler, Oded Schramm and Wendelin Werner proved that exact solutions to two problems in Brownian motion can be found without bending the rules. It earned them a Fields medal, the maths equivalent of a Nobel prize. More recently, Stanislav Smirnov at the University of Geneva in Switzerland solved a related problem, which resulted in him being awarded the Fields medal in 2010.

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4. Impossibility theorems

There are plenty of mathematical expressions that have no exact solution. Take one of the most famous numbers ever, pi, which is the ratio of a circle’s circumference to its diameter. Proving that it was impossible for pi’s digits after the decimal point to ever end was one of the greatest contributions to maths. Physicists similarly say that it is impossible to find solutions to certain problems, like finding the exact energies of electrons orbiting a helium atom. But can we prove that impossibility?

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5. Spin glass

To understand this problem, you need to know about spin, a quantum mechanical property of atoms and particles like electrons, which underlies magnetism. You can think of it like an arrow that can point up or down. Electrons inside blocks of materials are happiest if they sit next to electrons that have the opposite spin, but there are some arrangements where that isn’t possible. In these frustrated magnets, spins often flip around randomly in a way that, it turns out, is a useful model of other disordered systems including financial markets. But we have limited ways of mathematically describing how systems like this behave. This spin glass question asks if we can find a good way of doing it.

• See the full list of unsolved problems:  Open Problems in Mathematical Physics  

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7 of the hardest math problems that have yet to be solved — part 1

In this two-part article we take a look at some of the hardest unsolved problems in maths. first, let’s look at the riemann hypothesis and the twin prime conjecture..

Tejasri Gururaj

gorodenkoff  

  • Math problems like the Poincaré conjecture and Fermat's last theorem took centuries to solve.
  • However, others like the Riemann hypothesis and Goldbach's conjecture still haunt mathematicians and inspire new generations to find solutions.
  • Here we take a look at some of the hardest unsolved problems in math.

Many math problems take mathematicians decades and centuries to solve, while others continue to defy solutions.

There are substantial financial rewards for anyone who can solve some of these unsolved problems, encouraging mathematicians and problem solvers to take up the challenge.

Stay ahead of your peers in technology and engineering - The Blueprint

In this two-part article, we take a look at some of the hardest mathematical problems that remain unsolved to this day. In this first part, we discuss seven of them, beginning with the Collatz conjecture.

Give it a crack; you might just end up solving them!

1. The Collatz conjecture

worlds hardest math problem solved

Hugo Spinelli  

The Collatz conjecture, sometimes referred to as the 3n+1 problem, stands as one of the most renowned unsolved puzzles in mathematics. It seeks to answer a seemingly simple question: Can a series of basic arithmetic operations transform any positive integer into 1? 

The process involves generating sequences of integers, with each term derived from the previous one according to two rules. If the preceding number is even, it is divided by 2 to get the next number in the sequence. If the preceding term is odd, the next number is calculated by tripling it and adding 1.

For example, if we start with the number 5, the next number would be 3x5+1, which is 16. Since 16 is even, we divide it by 2 to get the next number in the sequence: 8. This continues until it reaches 1. 

So, this sequence would be: 5, 16, 8, 4, 2, 1

The central question is whether this sequence will always reach 1, irrespective of the starting positive integer.

The Collatz conjecture was introduced by the German mathematician Lothar Collatz in 1937, just two years after he had earned his doctorate. 

worlds hardest math problem solved

Over the years, many mathematicians have attempted to unravel the mystery of this conjecture, but it has remained an enigma. Many mathematicians have suggested that this problem may even be out of the reach of present-day mathematics. 

Despite the numerous efforts invested in exploring this conjecture, it remains unsolved, and the mathematical community continues to grapple with its intricacies.

2. The Goldbach conjecture

worlds hardest math problem solved

Christian Goldbach  

The Goldbach conjecture is one of the most famous unresolved questions in number theory and mathematics. It suggests that every even natural number greater than 2 can be represented as the sum of two prime numbers .

For example: 16 = 3 + 13

The conjecture was first proposed by Christian Goldbach in a letter to Leonhard Euler on June 7, 1742. In this letter, Goldbach presented the idea and conjectured that every integer greater than 2 could be expressed as the sum of three primes. 

Euler, in his reply, found the first part of Goldbach's conjecture highly probable, stating that "every even integer is a sum of two primes," although he couldn't provide a proof.

Over the years, significant progress has been made towards understanding the conjecture. For instance, Nils Pipping verified the statement up to n = 100,000 in 1938, and later, with the advent of computers, T. Oliveira e Silva ran a distributed computer search that confirmed the conjecture for n less than or equal to 4×10 18 (and double-checked up to 4×10 17 ) by 2013.

However, a complete, rigorous proof for all even integers greater than 2 remains elusive.

3. The twin prime conjecture

The twin prime conjecture revolves, unsurprisingly enough, around twin primes. These are prime numbers that are either 2 less or 2 more than another prime number, forming prime pairs like (3, 5) or (17, 19). 

The conjecture states that there are an infinite number of primes p such that p + 2 is also primes.

Proving the twin prime conjecture has been a long-standing open question in number theory. It was initially proposed by de Polignac in 1849, suggesting that for every natural number k, there are infinitely many primes p such that p+2k is also prime. The case k = 1 is the specific twin prime conjecture we are focusing on.

In 2013, work by Yitang Zhang marked a significant step toward proving the existence of an infinite number of twin primes. His research showed that there is a finite upper bound––70 million––for which the gaps between pairs of primes persist infinitely often. 

By April 2014, the bound (difference between the two primes) had been reduced to 246, indicating significant progress in understanding twin primes.

4. The Riemann hypothesis

worlds hardest math problem solved

Wikimedia Commons  

The Riemann hypothesis deals with the behavior of the Riemann zeta function, a mathematical function used to study the distribution of prime numbers in the complex plane.

It posits that all nontrivial zeros of the Riemann zeta function lie on a particular critical line in the complex plane. We can think of this critical line as like a tightrope, and the question is whether all these zeros gracefully balance on this line.

The story of the Riemann hypothesis starts in the 19th century with German mathematician Bernhard Riemann. In 1859, Riemann published a seminal paper titled " On the Number of Primes Less Than a Given Magnitude ." 

Within the pages of this document, he introduced the zeta function ζ(s), a complex function of a complex variable 's.' Riemann's revelation was revolutionary because it held the potential to unlock the secrets of prime numbers.

The Riemann hypothesis carries even greater weight in the mathematical world due to its inclusion among the prestigious "Millennium Prize Problems." 

In 2000, the Clay Mathematics Institute recognized the conjecture as one of seven outstanding mathematical challenges and offered a million-dollar prize for anyone who could furnish a correct proof for one of the seven. The essence of the Riemann hypothesis's significance lies in its connection to the prime number theorem.

5. The existence of odd perfect numbers

worlds hardest math problem solved

The existence of odd perfect numbers is a profound and unsolved mathematical mystery. Mathematically, a perfect number is a positive integer 'n' that is equal to the sum of its proper divisors, excluding the number itself. In other words:

n = 1 + 2 + 3 + ... + (n-1)

A well-known example of a perfect number is 28, with divisors 28 are 1, 2, 4, 7, and 14.

1 + 2 + 4 + 7 + 14 = 28.

However, it remains uncertain whether any odd perfect numbers exist.

In 1496, Jacques Lefèvre proposed the idea that all perfect numbers could be generated following Euclid's rule. This implied that no odd perfect number could exist, setting the stage for centuries of speculation.

Euler acknowledged the challenge by stating, "Whether... there are any odd perfect numbers is a most difficult question." 

In more recent times, Carl Pomerance presented a heuristic argument suggesting that the existence of odd perfect numbers is highly unlikely. This argument adds to the skepticism surrounding their existence.

6. Are these transcendental?

worlds hardest math problem solved

Wikipedia  

The problem at hand revolves around the transcendental nature of certain mathematical constants, specifically the Euler-Mascheroni Constant (γ) and the sum of π (pi) and e (Euler's number). 

A transcendental number is one that is not a root of any non-zero polynomial equation with integer coefficients. In simpler terms, transcendental numbers cannot be expressed as the solution to a polynomial equation where the coefficients are integers.

The Euler-Mascheroni constant, denoted as γ, is a fundamental mathematical constant that arises in various areas of mathematics, including number theory and calculus. 

Its transcendence status has been a matter of conjecture for years. Although there is substantial evidence pointing to the transcendental nature of γ, rigorous proof remains elusive.

Similarly, the sum of π and e is another example. Both π and e are transcendental numbers, and their sum is expected to be transcendental as well. However, this conjecture has not yet been definitively proven, adding an element of mystery to the mathematical world.

The status of these constants as transcendental or not transcends the boundaries of mathematical curiosity and holds significance in various mathematical disciplines.

7. The solitary number problem

worlds hardest math problem solved

The solitary number problem delves into the realm of solitary numbers, which are integers that don't have any 'friends; in the mathematical sense (e.g., they don't share a common relationship with any other numbers). Friendly numbers are those which have the same abundancy index (the ratio of the sum of divisors of a number to the number itself).

Solitary numbers include prime numbers, prime powers, and those numbers for which the greatest common divisor of the number and the sum of its divisors (denoted as sigma(n)) equals 1.

For example, the number 5 is a solitary number.  The divisors of 5 are 1 and 5, and their sum is 6. The greatest common divisor of 5 and 6 is 1.

While some numbers can be proven to be solitary by examining their properties, there are others for which proving solitariness is challenging. For instance, it's believed that numbers like 10, 15, 20, and many more are solitary, but providing conclusive proof remains elusive.

The concept of solitary numbers has intrigued mathematicians for years. While prime numbers are well-known solitary numbers, other integers also exhibit solitariness, even when they don't share the greatest common divisor of 1 with sigma(n).

In 1997, mathematician Carl Pomerance made an intriguing claim that solitary numbers possess positive density, a measure of how frequently certain elements or objects occur within a given set or population.

This challenged a 1977 conjecture by Anderson and Hickerson; however, this proof was never published and remains elusive, casting uncertainty over the claim's status. Classifying numbers as friendly or solitary poses a formidable challenge in number theory.

Consider the number 10, among the smallest with an unknown classification, conjectured to be solitary. If it is not solitary, its smallest friend is estimated to be an incredibly large number, on the order of 10 30 .

In 2022, Sourav Mandal shed light on the potential nature of 10's friend, proposing a specific form it must follow if it exists, adding an intriguing layer to the problem. Furthermore, examples like 24, classified as friendly, and possessing 91,963,648 as its smallest friend, illustrate the diversity in the classification of numbers as friendly or solitary.

And that’s done for part 1! Don’t forget to check out part 2 for seven more unsolved problems in mathematics that are challenging mathematicians to think outside the box.

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Sat / act prep online guides and tips, the 15 hardest sat math questions ever.

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Want to test yourself against the most difficult SAT math questions? Want to know what makes these questions so difficult and how best to solve them? If you're ready to really sink your teeth into the SAT math section and have your sights set on that perfect score, then this is the guide for you.

We've put together what we believe to be the 15 most difficult questions for the current SAT , with strategies and answer explanations for each. These are all hard SAT Math questions from College Board SAT practice tests, which means understanding them is one of the best ways to study for those of you aiming for perfection.

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Brief Overview of SAT Math

The third and fourth sections of the SAT will always be math sections . The first math subsection (labeled "3") does not allow you to use a calculator, while the second math subsection (labeled as "4") does allow the use of a calculator. Don't worry too much about the no-calculator section, though: if you're not allowed to use a calculator on a question, it means you don't need a calculator to answer it.

Each math subsection is arranged in order of ascending difficulty (where the longer it takes to solve a problem and the fewer people who answer it correctly, the more difficult it is). On each subsection, question 1 will be "easy" and question 15 will be considered "difficult." However, the ascending difficulty resets from easy to hard on the grid-ins.

Hence, multiple choice questions are arranged in increasing difficulty (questions 1 and 2 will be the easiest, questions 14 and 15 will be the hardest), but the difficulty level resets for the grid-in section (meaning questions 16 and 17 will again be "easy" and questions 19 and 20 will be very difficult).

With very few exceptions, then, the most difficult SAT math problems will be clustered at the end of the multiple choice segments or the second half of the grid-in questions. In addition to their placement on the test, though, these questions also share a few other commonalities. In a minute, we'll look at example questions and how to solve them, then analyze them to figure out what these types of questions have in common.

But First: Should You Be Focusing on the Hardest Math Questions Right Now?

If you're just getting started in your study prep (or if you've simply skipped this first, crucial step), definitely stop and take a full practice test to gauge your current scoring level. Check out our guide to all the free SAT practice tests available online and then sit down to take a test all at once.

The absolute best way to assess your current level is to simply take the SAT practice test as if it were real , keeping strict timing and working straight through with only the allowed breaks (we know—probably not your favorite way to spend a Saturday). Once you've got a good idea of your current level and percentile ranking, you can set milestones and goals for your ultimate SAT Math score.

If you're currently scoring in the 200-400 or the 400-600 range on SAT Math, your best bet is first to check out our guide to improving your math score to be consistently at or over a 600 before you start in trying to tackle the most difficult math problems on the test.

If, however, you're already scoring above a 600 on the Math section and want to test your mettle for the real SAT, then definitely proceed to the rest of this guide. If you're aiming for perfect (or close to) , then you'll need to know what the most difficult SAT math questions look like and how to solve them. And luckily, that's exactly what we'll do.

WARNING: Since there are a limited number of official SAT practice tests , you may want to wait to read this article until you've attempted all or most of the first four official practice tests (since most of the questions below were taken from those tests). If you're worried about spoiling those tests, stop reading this guide now; come back and read it when you've completed them.

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Now let's get to our list of questions (whoo)!

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The 15 Hardest SAT Math Questions

Now that you're sure you should be attempting these questions, let's dive right in! We've curated 15 of the most difficult SAT Math questions for you to try below, along with walkthroughs of how to get the answer (if you're stumped).

No Calculator SAT Math Questions

$$C=5/9(F-32)$$

The equation above shows how temperature $F$, measured in degrees Fahrenheit, relates to a temperature $C$, measured in degrees Celsius. Based on the equation, which of the following must be true?

  • A temperature increase of 1 degree Fahrenheit is equivalent to a temperature increase of $5/9$ degree Celsius.
  • A temperature increase of 1 degree Celsius is equivalent to a temperature increase of 1.8 degrees Fahrenheit.
  • A temperature increase of $5/9$ degree Fahrenheit is equivalent to a temperature increase of 1 degree Celsius.

A) I only B) II only C) III only D) I and II only

ANSWER EXPLANATION: Think of the equation as an equation for a line

where in this case

$$C= {5}/{9} (F−32)$$

$$C={5}/{9}F −{5}/{9}(32)$$

You can see the slope of the graph is ${5}/{9}$, which means that for an increase of 1 degree Fahrenheit, the increase is ${5}/{9}$ of 1 degree Celsius.

$$C= {5}/{9} (F)$$

$$C= {5}/{9} (1)= {5}/{9}$$

Therefore, statement I is true. This is the equivalent to saying that an increase of 1 degree Celsius is equal to an increase of ${9}/{5}$ degrees Fahrenheit.

$$1= {5}/{9} (F)$$

$$(F)={9}/{5}$$

Since ${9}/{5}$ = 1.8, statement II is true.

The only answer that has both statement I and statement II as true is D , but if you have time and want to be absolutely thorough, you can also check to see if statement III (an increase of ${5}/{9}$ degree Fahrenheit is equal to a temperature increase of 1 degree Celsius) is true:

$$C= {5}/{9} ({5}/{9})$$

$$C= {25} /{81} (\which \is ≠ 1)$$

An increase of $5/9$ degree Fahrenheit leads to an increase of ${25}/{81}$, not 1 degree, Celsius, and so Statement III is not true.

The final answer is D.

The equation ${24x^2 + 25x -47}/{ax-2} = -8x-3-{53/{ax-2}}$ is true for all values of $x≠2/a$, where $a$ is a constant.

What is the value of $a$?

A) -16 B) -3 C) 3 D) 16

ANSWER EXPLANATION: There are two ways to solve this question. The faster way is to multiply each side of the given equation by $ax-2$ (so you can get rid of the fraction). When you multiply each side by $ax-2$, you should have:

$$24x^2 + 25x - 47 = (-8x-3)(ax-2) - 53$$

You should then multiply $(-8x-3)$ and $(ax-2)$ using FOIL.

$$24x^2 + 25x - 47 = -8ax^2 - 3ax +16x + 6 - 53$$

Then, reduce on the right side of the equation

$$24x^2 + 25x - 47 = -8ax^2 - 3ax +16x - 47$$

Since the coefficients of the $x^2$-term have to be equal on both sides of the equation, $−8a = 24$, or $a = −3$.

The other option which is longer and more tedious is to attempt to plug in all of the answer choices for a and see which answer choice makes both sides of the equation equal. Again, this is the longer option, and I do not recommend it for the actual SAT as it will waste too much time.

The final answer is B.

If $3x-y = 12$, what is the value of ${8^x}/{2^y}$?

A) $2^{12}$ B) $4^4$ C) $8^2$ D) The value cannot be determined from the information given.

ANSWER EXPLANATION: One approach is to express

$${8^x}/{2^y}$$

so that the numerator and denominator are expressed with the same base. Since 2 and 8 are both powers of 2, substituting $2^3$ for 8 in the numerator of ${8^x}/{2^y}$ gives

$${(2^3)^x}/{2^y}$$

which can be rewritten

$${2^3x}/{2^y}$$

Since the numerator and denominator of have a common base, this expression can be rewritten as $2^(3x−y)$. In the question, it states that $3x − y = 12$, so one can substitute 12 for the exponent, $3x − y$, which means that

$${8^x}/{2^y}= 2^12$$

The final answer is A.

Points A and B lie on a circle with radius 1, and arc ${AB}↖⌢$ has a length of $π/3$. What fraction of the circumference of the circle is the length of arc ${AB}↖⌢$?

ANSWER EXPLANATION: To figure out the answer to this question, you'll first need to know the formula for finding the circumference of a circle.

The circumference, $C$, of a circle is $C = 2πr$, where $r$ is the radius of the circle. For the given circle with a radius of 1, the circumference is $C = 2(π)(1)$, or $C = 2π$.

To find what fraction of the circumference the length of ${AB}↖⌢$ is, divide the length of the arc by the circumference, which gives $π/3 ÷ 2π$. This division can be represented by $π/3 * {1/2}π = 1/6$.

The fraction $1/6$ can also be rewritten as $0.166$ or $0.167$.

The final answer is $1/6$, $0.166$, or $0.167$.

$${8-i}/{3-2i}$$

If the expression above is rewritten in the form $a+bi$, where $a$ and $b$ are real numbers, what is the value of $a$? (Note: $i=√{-1}$)

ANSWER EXPLANATION: To rewrite ${8-i}/{3-2i}$ in the standard form $a + bi$, you need to multiply the numerator and denominator of ${8-i}/{3-2i}$ by the conjugate, $3 + 2i$. This equals

$$({8-i}/{3-2i})({3+2i}/{3+2i})={24+16i-3+(-i)(2i)}/{(3^2)-(2i)^2}$$

Since $i^2=-1$, this last fraction can be reduced simplified to

$$ {24+16i-3i+2}/{9-(-4)}={26+13i}/{13}$$

which simplifies further to $2 + i$. Therefore, when ${8-i}/{3-2i}$ is rewritten in the standard form a + bi, the value of a is 2.

In triangle $ABC$, the measure of $∠B$ is 90°, $BC=16$, and $AC$=20. Triangle $DEF$ is similar to triangle $ABC$, where vertices $D$, $E$, and $F$ correspond to vertices $A$, $B$, and $C$, respectively, and each side of triangle $DEF$ is $1/3$ the length of the corresponding side of triangle $ABC$. What is the value of $sinF$?

ANSWER EXPLANATION: Triangle ABC is a right triangle with its right angle at B. Therefore, $\ov {AC}$ is the hypotenuse of right triangle ABC, and $\ov {AB}$ and $\ov {BC}$ are the legs of right triangle ABC. According to the Pythagorean theorem,

$$AB =√{20^2-16^2}=√{400-256}=√{144}=12$$

Since triangle DEF is similar to triangle ABC, with vertex F corresponding to vertex C, the measure of $\angle ∠ {F}$ equals the measure of $\angle ∠ {C}$. Therefore, $sin F = sin C$. From the side lengths of triangle ABC,

$$sinF ={\opposite \side}/{\hypotenuse}={AB}/{AC}={12}/{20}={3}/{5}$$

Therefore, $sinF ={3}/{5}$.

The final answer is ${3}/{5}$ or 0.6.

Calculator-Allowed SAT Math Questions

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The incomplete table above summarizes the number of left-handed students and right-handed students by gender for the eighth grade students at Keisel Middle School. There are 5 times as many right-handed female students as there are left-handed female students, and there are 9 times as many right-handed male students as there are left-handed male students. if there is a total of 18 left-handed students and 122 right-handed students in the school, which of the following is closest to the probability that a right-handed student selected at random is female? (Note: Assume that none of the eighth-grade students are both right-handed and left-handed.)

A) 0.410 B) 0.357 C) 0.333 D) 0.250

ANSWER EXPLANATION: In order to solve this problem, you should create two equations using two variables ($x$ and $y$) and the information you're given. Let $x$ be the number of left-handed female students and let $y$ be the number of left-handed male students. Using the information given in the problem, the number of right-handed female students will be $5x$ and the number of right-handed male students will be $9y$. Since the total number of left-handed students is 18 and the total number of right-handed students is 122, the system of equations below must be true:

$$x + y = 18$$

$$5x + 9y = 122$$

When you solve this system of equations, you get $x = 10$ and $y = 8$. Thus, 5*10, or 50, of the 122 right-handed students are female. Therefore, the probability that a right-handed student selected at random is female is ${50}/{122}$, which to the nearest thousandth is 0.410.

Questions 8 & 9

Use the following information for both question 7 and question 8.

If shoppers enter a store at an average rate of $r$ shoppers per minute and each stays in the store for average time of $T$ minutes, the average number of shoppers in the store, $N$, at any one time is given by the formula $N=rT$. This relationship is known as Little's law.

The owner of the Good Deals Store estimates that during business hours, an average of 3 shoppers per minute enter the store and that each of them stays an average of 15 minutes. The store owner uses Little's law to estimate that there are 45 shoppers in the store at any time.

Little's law can be applied to any part of the store, such as a particular department or the checkout lines. The store owner determines that, during business hours, approximately 84 shoppers per hour make a purchase and each of these shoppers spend an average of 5 minutes in the checkout line. At any time during business hours, about how many shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store?

ANSWER EXPLANATION: Since the question states that Little's law can be applied to any single part of the store (for example, just the checkout line), then the average number of shoppers, $N$, in the checkout line at any time is $N = rT$, where $r$ is the number of shoppers entering the checkout line per minute and $T$ is the average number of minutes each shopper spends in the checkout line.

Since 84 shoppers per hour make a purchase, 84 shoppers per hour enter the checkout line. However, this needs to be converted to the number of shoppers per minute (in order to be used with $T = 5$). Since there are 60 minutes in one hour, the rate is ${84 \shoppers \per \hour}/{60 \minutes} = 1.4$ shoppers per minute. Using the given formula with $r = 1.4$ and $T = 5$ yields

$$N = rt = (1.4)(5) = 7$$

Therefore, the average number of shoppers, $N$, in the checkout line at any time during business hours is 7.

The final answer is 7.

The owner of the Good Deals Store opens a new store across town. For the new store, the owner estimates that, during business hours, an average of 90 shoppers per hour enter the store and each of them stays an average of 12 minutes. The average number of shoppers in the new store at any time is what percent less than the average number of shoppers in the original store at any time? (Note: Ignore the percent symbol when entering your answer. For example, if the answer is 42.1%, enter 42.1)

ANSWER EXPLANATION: According to the original information given, the estimated average number of shoppers in the original store at any time (N) is 45. In the question, it states that, in the new store, the manager estimates that an average of 90 shoppers per hour (60 minutes) enter the store, which is equivalent to 1.5 shoppers per minute (r). The manager also estimates that each shopper stays in the store for an average of 12 minutes (T). Thus, by Little's law, there are, on average, $N = rT = (1.5)(12) = 18$ shoppers in the new store at any time. This is

$${45-18}/{45} * 100 = 60$$

percent less than the average number of shoppers in the original store at any time.

The final answer is 60.

Question 10

In the $xy$-plane, the point $(p,r)$ lies on the line with equation $y=x+b$, where $b$ is a constant. The point with coordinates $(2p, 5r)$ lies on the line with equation $y=2x+b$. If $p≠0$, what is the value of $r/p$?

ANSWER EXPLANATION: Since the point $(p,r)$ lies on the line with equation $y=x+b$, the point must satisfy the equation. Substituting $p$ for $x$ and $r$ for $y$ in the equation $y=x+b$ gives $r=p+b$, or $\bi b$ = $\bi r-\bi p$.

Similarly, since the point $(2p,5r)$ lies on the line with the equation $y=2x+b$, the point must satisfy the equation. Substituting $2p$ for $x$ and $5r$ for $y$ in the equation $y=2x+b$ gives:

$5r=2(2p)+b$

$\bi b$ = $\bo 5 \bi r-\bo 4\bi p$.

Next, we can set the two equations equal to $b$ equal to each other and simplify:

$b=r-p=5r-4p$

Finally, to find $r/p$, we need to divide both sides of the equation by $p$ and by $4$:

The correct answer is B , $3/4$.

If you picked choices A and D, you may have incorrectly formed your answer out of the coefficients in the point $(2p, 5r)$. If you picked Choice C, you may have confused $r$ and $p$.

Note that while this is in the calculator section of the SAT, you absolutely do not need your calculator to solve it!

Question 11

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A) 261.8 B) 785.4 C) 916.3 D) 1047.2

ANSWER EXPLANATION: The volume of the grain silo can be found by adding the volumes of all the solids of which it is composed (a cylinder and two cones). The silo is made up of a cylinder (with height 10 feet and base radius 5 feet) and two cones (each with height 5 ft and base radius 5 ft). The formulas given at the beginning of the SAT Math section:

Volume of a Cone

$$V={1}/{3}πr^2h$$

Volume of a Cylinder

$$V=πr^2h$$

can be used to determine the total volume of the silo. Since the two cones have identical dimensions, the total volume, in cubic feet, of the silo is given by

$$V_{silo}=π(5^2)(10)+(2)({1}/{3})π(5^2)(5)=({4}/{3})(250)π$$

which is approximately equal to 1,047.2 cubic feet.

Question 12

If $x$ is the average (arithmetic mean) of $m$ and $9$, $y$ is the average of $2m$ and $15$, and $z$ is the average of $3m$ and $18$, what is the average of $x$, $y$, and $z$ in terms of $m$?

A) $m+6$ B) $m+7$ C) $2m+14$ D) $3m + 21$

ANSWER EXPLANATION: Since the average (arithmetic mean) of two numbers is equal to the sum of the two numbers divided by 2, the equations $x={m+9}/{2}$, $y={2m+15}/{2}$, $z={3m+18}/{2}$are true. The average of $x$, $y$, and $z$ is given by ${x + y + z}/{3}$. Substituting the expressions in m for each variable ($x$, $y$, $z$) gives

$$[{m+9}/{2}+{2m+15}/{2}+{3m+18}/{2}]/3$$

This fraction can be simplified to $m + 7$.

Question 13

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The function $f(x)=x^3-x^2-x-{11/4}$ is graphed in the $xy$-plane above. If $k$ is a constant such that the equation $f(x)=k$ has three real solutions, which of the following could be the value of $k$?

ANSWER EXPLANATION: The equation $f(x) = k$ gives the solutions to the system of equations

$$y = f(x) = x^3-x^2-x-{11}/{4}$$

A real solution of a system of two equations corresponds to a point of intersection of the graphs of the two equations in the $xy$-plane.

The graph of $y = k$ is a horizontal line that contains the point $(0, k)$ and intersects the graph of the cubic equation three times (since it has three real solutions). Given the graph, the only horizontal line that would intersect the cubic equation three times is the line with the equation $y = −3$, or $f(x) = −3$. Therefore, $k$ is $-3$.

Question 14

$$q={1/2}nv^2$$

The dynamic pressure $q$ generated by a fluid moving with velocity $v$ can be found using the formula above, where $n$ is the constant density of the fluid. An aeronautical engineer users the formula to find the dynamic pressure of a fluid moving with velocity $v$ and the same fluid moving with velocity 1.5$v$. What is the ratio of the dynamic pressure of the faster fluid to the dynamic pressure of the slower fluid?

ANSWER EXPLANATION: To solve this problem, you need to set up to equations with variables. Let $q_1$ be the dynamic pressure of the slower fluid moving with velocity $v_1$, and let $q_2$ be the dynamic pressure of the faster fluid moving with velocity $v_2$. Then

$$v_2 =1.5v_1$$

Given the equation $q = {1}/{2}nv^2$, substituting the dynamic pressure and velocity of the faster fluid gives $q_2 = {1}/{2}n(v_2)^2$. Since $v_2 =1.5v_1$, the expression $1.5v_1$ can be substituted for $v_2$ in this equation, giving $q_2 = {1}/{2}n(1.5v_1)^2$. By squaring $1.5$, you can rewrite the previous equation as

$$q_2 = (2.25)({1}/{2})n(v_1)^2 = (2.25)q_1$$

Therefore, the ratio of the dynamic pressure of the faster fluid is

$${q2}/{q1} = {2.25 q_1}/{q_1}= 2.25$$

The final answer is 2.25 or 9/4.

Question 15

For a polynomial $p(x)$, the value of $p(3)$ is $-2$. Which of the following must be true about $p(x)$?

A) $x-5$ is a factor of $p(x)$. B) $x-2$ is a factor of $p(x)$. C) $x+2$ is a factor of $p(x)$. D) The remainder when $p(x)$ is divided by $x-3$ is $-2$.

ANSWER EXPLANATION: If the polynomial $p(x)$ is divided by a polynomial of the form $x+k$ (which accounts for all of the possible answer choices in this question), the result can be written as

$${p(x)}/{x+k}=q(x)+{r}/{x+k}$$

where $q(x)$ is a polynomial and $r$ is the remainder. Since $x + k$ is a degree-1 polynomial (meaning it only includes $x^1$ and no higher exponents), the remainder is a real number.

Therefore, $p(x)$ can be rewritten as $p(x) = (x + k)q(x) + r$, where $r$ is a real number.

The question states that $p(3) = -2$, so it must be true that

$$-2 = p(3) = (3 + k)q(3) + r$$

Now we can plug in all the possible answers. If the answer is A, B, or C, $r$ will be $0$, while if the answer is D, $r$ will be $-2$.

A. $-2 = p(3) = (3 + (-5))q(3) + 0$ $-2=(3-5)q(3)$ $-2=(-2)q(3)$

This could be true, but only if $q(3)=1$

B. $-2 = p(3) = (3 + (-2))q(3) + 0$ $-2 = (3-2)q(3)$ $-2 = (-1)q(3)$

This could be true, but only if $q(3)=2$

C. $-2 = p(3) = (3 + 2)q(3) + 0$ $-2 = (5)q(3)$

This could be true, but only if $q(3)={-2}/{5}$

D. $-2 = p(3) = (3 + (-3))q(3) + (-2)$ $-2 = (3 - 3)q(3) + (-2)$ $-2 = (0)q(3) + (-2)$

This will always be true no matter what $q(3)$ is.

Of the answer choices, the only one that must be true about $p(x)$ is D, that the remainder when $p(x)$ is divided by $x-3$ is -2.

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You deserve all the naps after running through those questions.

What Do the Hardest SAT Math Questions Have in Common?

It's important to understand what makes these hard questions "hard." By doing so, you'll be able to both understand and solve similar questions when you see them on test day, as well as have a better strategy for identifying and correcting your previous SAT math errors.

In this section, we'll look at what these questions have in common and give examples of each type. Some of the reasons why the hardest math questions are the hardest math questions is because they:

#1: Test Several Mathematical Concepts at Once

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Here, we must deal with imaginary numbers and fractions all at once.

Secret to success: Think of what applicable math you could use to solve the problem, do one step at a time, and try each technique until you find one that works!

#2: Involve a Lot of Steps

Remember: the more steps you need to take, the easier to mess up somewhere along the line!

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We must solve this problem in steps (doing several averages) to unlock the rest of the answers in a domino effect. This can get confusing, especially if you're stressed or running out of time.

Secret to success: Take it slow, take it step by step, and double-check your work so you don't make mistakes!

#3: Test Concepts That You Have Limited Familiarity With

For example, many students are less familiar with functions than they are with fractions and percentages, so most function questions are considered "high difficulty" problems.

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If you don't know your way around functions , this would be a tricky problem.

Secret to success: Review math concepts that you don't have as much familiarity with such as functions . We suggest using our great free SAT Math review guides .

#4: Are Worded in Unusual or Convoluted Ways

It can be difficult to figure out exactly what some questions are asking , much less figure out how to solve them. This is especially true when the question is located at the end of the section, and you are running out of time.

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Because this question provides so much information without a diagram, it can be difficult to puzzle through in the limited time allowed.

Secret to success: Take your time, analyze what is being asked of you, and draw a diagram if it's helpful to you.

#5: Use Many Different Variables

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With so many different variables in play, it is quite easy to get confused.

Secret to success: Take your time, analyze what is being asked of you, and consider if plugging in numbers is a good strategy to solve the problem (it wouldn't be for the question above, but would be for many other SAT variable questions).

The Take-Aways

The SAT is a marathon and the better prepared you are for it, the better you'll feel on test day. Knowing how to handle the hardest questions the test can throw at you will make taking the real SAT seem a lot less daunting.

If you felt that these questions were easy, make sure not underestimate the effect of adrenaline and fatigue on your ability to solve problems. As you continue to study, always adhere to the proper timing guidelines and try to take full tests whenever possible. This is the best way to recreate the actual testing environment so that you can prepare for the real deal.

If you felt these questions were challenging, be sure to strengthen your math knowledge by checking out our individual math topic guides for the SAT . There, you'll see more detailed explanations of the topics in question as well as more detailed answer breakdowns.

What's Next?

Felt that these questions were harder than you were expecting? Take a look at all the topics covered in the SAT math section and then note which sections were particular difficulty for you. Next, take a gander at our individual math guides to help you shore up any of those weak areas.

Running out of time on the SAT math section? Our guide will help you beat the clock and maximize your score .

Aiming for a perfect score? Check out our guide on how to get a perfect 800 on the SAT math section , written by a perfect-scorer.

Want to improve your SAT score by 160 points?

Check out our best-in-class online SAT prep classes . We guarantee your money back if you don't improve your SAT score by 160 points or more.

Our classes are entirely online, and they're taught by SAT experts . If you liked this article, you'll love our classes. Along with expert-led classes, you'll get personalized homework with thousands of practice problems organized by individual skills so you learn most effectively. We'll also give you a step-by-step, custom program to follow so you'll never be confused about what to study next.

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Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.

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9 Hardest Calculus Problems Ever You’ll Ever Encounter

Hardest Calculus Problems Ever

Lately, I was teaching one of the brightest students; she asked me what the hardest calculus problem ever was. Her question led me to do deeper research to find.

Mathematics is a constantly evolving field, and new equations and calculations are constantly being discovered. But some problems have posed a challenge for mathematicians for centuries. Here are the

In this blog, I wanted to share what I found; maybe you try solving it. Read on to find some of the most challenging calculus problems and discuss why they are so difficult.

The good news is two of the hardest calculus problems are still unsolved, and there is a reward of $1 million dollars for whoever finds the answers to each problem.

You might also enjoy reading: What Jobs Can You Get With a Mathematics Degree: 9 Best Options .

Table of Contents

1- The Three-Body Problem

The Three-Body Problem is one of the oldest and most famous unsolved problems in mathematics. It was first proposed by Isaac Newton in 1687 and remains unsolved to this day (Source: Scientific American )

The three-body problem deals with understanding the motion of three objects interacting with each other, such as moons orbiting planets or stars in galaxies, given their initial positions and velocities.

It has been particularly difficult for mathematicians due to its chaotic behavior, meaning that small changes in the initial conditions can lead to drastically different outcomes. Additionally, its nonlinearity makes it resistant to traditional mathematical techniques.

When Isaac Newton published his  Principia  in 1687, he asked: “How will two masses move in space if the only force on them is their mutual gravitational attraction? ” Newton formulated the question as a problem solving a system of differential equations . 

Despite these challenges, many researchers have made significant progress on the Three-Body Problem over the years, but it still remains unsolved (Source: Popular Mechanics )

Watch the video below to learn more about the Three-Body Problem.

2- Goldbach’s Conjecture

Christian Goldbach first proposed this conjecture in 1742 and stated that every even number greater than two could be written as the sum of two prime numbers (a prime number is an integer greater than one with no divisors other than itself).

For example, 8 = 3 + 5 or 10 = 7 + 3. While this conjecture seems simple enough at first glance, it has proven surprisingly hard to prove or disprove!

Despite intense effort from mathematicians worldwide over 250 years, Goldbach’s Conjecture remains unproven and stands as one of the greatest open problems in mathematics today.

Further progress on Goldbach’s conjecture emerged in 1973 when the Chinese mathematician Chen Jing Run demonstrated that every sufficiently large even number is the sum of a prime and a number with at most two prime factors .

Hardest Calculus Problem Ever

3- Fermat’s Last Theorem

This theorem dates back to 1637 when Pierre de Fermat wrote down his famous equation without providing any proof or explanation for it in his notebook: “it is impossible to separate a cube into two cubes or a fourth power into two fourth powers or generally any power higher than second into two like powers.”

Fermat’s last theorem, also known as Fermat’s great theorem, is the statement that there exist no natural numbers (1, 2, 3,…) x, y, and z such that x^n + y^n = z^n, in which n is a natural number bigger than 2.

 For instance, if  n  = 3, Fermat’s last theorem says that no natural numbers  x ,  y , and  z  exist such that  x^ 3 +  y  ^3 =  z^ 3. In other words, the sum of two cubes is not a cube (Source: Britannica )

The Fermat’s Last Theorem remained unproven until 1995 when Andrew Wiles finally provided proof using elliptic curves after working on it for seven years (Source: National Science Foundation (NSF) )

This theorem stands as one of the greatest achievements in mathematics and still remains one of the most difficult problems ever tackled by mathematicians worldwide.
Feynman wrote an unpublished 2 page manuscript approaching Fermat’s Last Theorem from a probabilistic standpoint and concluded (before Andrew Wiles’ proof!) that “for my money Fermat’s theorem is true”. Here is the reconstruction of his approach: https://t.co/3GrUNXEfuW pic.twitter.com/sDpUD5JWJF — Fermat’s Library (@fermatslibrary) November 5, 2018

4- The Riemann Hypothesis

The Riemann Hypothesis is perhaps one of the most famous unsolved problems in mathematics today . It states that all non-trivial zeros of the Riemann zeta function have real parts equal to 1/2.

While it has not yet been proven (or disproven), mathematicians have made considerable progress towards solving it using techniques from complex analysis and number theory.

Unfortunately, many mathematicians believe it may never be solved without major mathematics and computer science breakthroughs due to its complexity and difficulty.

If you are looking for ways to make a million dollars by solving math, try solving the Riemann Hypothesis. It is among the  Seven Millennium Prize Problems , with a $1 million reward if you find its solutions.  

If you solve the Riemann Hypothesis tomorrow, it will open an avalanche of further progress. It would be massive news throughout the topics of Number Theory and Analysis. 

I suggest you watch the video below to learn more about the Riemann Hypothesis.

5- The Collatz Conjecture

The Collatz Conjecture is another unsolved mathematical problem that has remained a mystery since its inception in 1937. Intuitively described, it deals with the sequence created by taking any number and, if it is even, dividing it by two, and if it is odd, multiplying by three and adding one .

Every cycle of this algorithm eventually converges to the same number: 1. So far, no one has been able to determine why this happens or why the Collatz Conjecture holds true for all natural numbers (positive integers from 1 to infinity).

This elusive problem has stumped mathematicians for decades and continues to draw researchers to try and solve this head-scratching conundrum.

Despite numerous attempts made to unravel its secrets, the Collatz Conjecture remains as enigmatic as ever, begging us to discover its mystery and open up new doors in the realm of mathematics.

If proved true, the Collatz Conjecture could provide major new insights into our understanding of mathematics and computing algorithms, leading to numerous potential applications.

Undoubtedly, whoever solves this hypothesis will have made one of the great discoveries in mathematics.

The video below discusses the Collatz Conjecture.

6- The Twin Prime Conjecture

In number theory, the Twin Prime conjecture , also known as Polignac’s conjecture, asserts that infinitely many twin primes, or pairs of primes, differ by 2.  As an illustration, 3 and 5, 5 and 7, 11 and 13, and 17 and 19 are considered twin primes. As numbers become larger, primes become less frequent, and twin primes are rarer still.

The Twin Prime Conjecture is an unsolved problem in mathematics that has stumped the best minds for centuries. If the conjecture is true, it will open up a whole new realm of prime numbers, providing new avenues for exploration and even potential applications in cryptography.

However, it has been difficult to prove due to the lack of general patterns for consecutive primes; any pattern made thus far is inconsistent and unreliable at best.

Despite this difficulty, mathematicians remain optimistic about uncovering the answer to this mystery–and when they do, it will surely be a monumental achievement!

I encourage you to watch the video below to learn more about the Twin Prime Conjecture.

7- The Birch and Swinnerton-Dyer Conjecture

The Birch and Swinnerton-Dyer Conjecture , a crucial unsolved mathematical problem in number theory, remains one of the greatest mysteries of our time. The Birch and Swinnerton-Dyer Conjecture is also among the six unsolved Millennium Prize Problems, meaning that if you solve it, you will be rewarded with one million dollars.

Originally conjectured in the 1960s, this idiosyncratic conjecture has captivated mathematicians ever since. While researchers have gained insight into related topics such as elliptic curves and modular forms, the true complexity of this conjecture has still eluded them.

An elliptic curve is a particular kind of function that can be written in this form y²=x³+ax+b. It turns out that these types of functions have specific properties that explore other math topics, such as Algebra and Number Theory.

As a result, researchers continue to explore new approaches and hope they can one day demonstrate their veracity. Undoubtedly, this intriguing yet tough problem will captivate mathematicians for years to come.

If you are interested in learning more about the Birch-Swinnerton-Dyer Conjecture, I encourage you to watch the video below.

8- The Kissing Number Problem

The Kissing Number Problem has stumped mathematicians for centuries. The problem involves finding the maximum number of equal-sized spheres that can touch one central sphere without overlapping or leaving any spaces between them (Source: Princeton University )

Initially thought to be a simple problem to solve, it is quite challenging to determine this ‘kissing number accurately.’ The answer varies depending on the dimension of the space – in two dimensions, or a flat surface, it’s only six, but in three dimensions, it is much larger and still debated today.

The Kissing Number Problem continues to baffle modern mathematicians, providing an interesting and complex challenge that could lead to countless scientific advances.

Watch the video below to learn more about the Kissing Number Problem.

9- The Unknotting Problem

The Unknotting Problem has fascinated mathematicians since discovering that the unknot is equivalent to a one-dimensional closed loop in three-dimensional space.

The Unknotting Problem simply asks if a particular knot can be undone without changing its form. For example, questions such as “which knots have the fewest crossings?” or “can all knots be unknotted?”. It was first described in 1904 by Greek mathematician Peter Guthrie Tait . It is still an open problem with no known general algorithm for efficiently deciding whether a knot can be untied to become just a circle.

As fascinating as this perplexing problem is to mathematicians, understanding and solving the Unknotting Problem could be essential for researchers hoping to apply mathematics to biology and chemistry, where twists and turns play an important role in the workings of molecules.

Check out A Journey From Elementary to Advanced Mathematics: The Unknotting Problem if you want to learn more.

Here is an interesting PPT presentation about the Unknotting Problem.

Hardest Calculus Problem Ever

What to read next:

  • Can You Do A Level Maths In 1 Year? (And How to Ace A Level Math in a Year!)
  • What Does a Level Math Course Cover?
  • Introduction to Logarithmic Functions .

Wrapping Up

Mathematicians have been tackling difficult mathematical problems since time immemorial with varying degrees of success; some have been solved, while others remain unsolved mysteries today.

From Goldbach’s Conjecture to Fermat’s Last Theorem, challenging mathematical problems continue to captivate mathematicians everywhere and provide them with fascinating puzzles to solve.

Whether you are a high school student studying calculus or a college student taking a higher-level math course, there is always something interesting waiting for you if you look for it.

I encourage you to give some of these hardest math problems a try. Who knows – maybe you will come up with an answer and be rewarded with a million dollars prize.

I am Altiné. I am the guy behind mathodics.com. When I am not teaching math, you can find me reading, running, biking, or doing anything that allows me to enjoy nature's beauty. I hope you find what you are looking for while visiting mathodics.com.

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10 Math Equations That Have Never Been Solved

By Kathleen Cantor, 10 Sep 2020

Mathematics has played a major role in so many life-altering inventions and theories. But there are still some math equations that have managed to elude even the greatest minds, like Einstein and Hawkins. Other equations, however, are simply too large to compute. So for whatever reason, these puzzling problems have never been solved. But what are they?

Like the rest of us, you're probably expecting some next-level difficulty in these mathematical problems. Surprisingly, that is not the case. Some of these equations are even based on elementary school concepts and are easily understandable - just unsolvable.

1. The Riemann Hypothesis

Equation: σ (n) ≤ Hn +ln (Hn)eHn

  • Where n is a positive integer
  • Hn is the n-th harmonic number
  • σ(n) is the sum of the positive integers divisible by n

For an instance, if n = 4 then σ(4)=1+2+4=7 and H4 = 1+1/2+1/3+1/4. Solve this equation to either prove or disprove the following inequality n≥1? Does it hold for all n≥1?

This problem is referred to as Lagarias’s Elementary Version of the Riemann Hypothesis and has a price of a million dollars offered by the  Clay Mathematics Foundation  for its solution.

2. The Collatz Conjecture

Equation: 3n+1

  • where n is a positive integer n/2
  • where n is a non-negative integer

Prove the answer end by cycling through 1,4,2,1,4,2,1,… if n is a positive integer. This is a repetitive process and you will repeat it with the new value of n you get. If your first n = 1 then your subsequent answers will be 1, 4, 2, 1, 4, 2, 1, 4… infinitely. And if n = 5 the answers will be 5,16,8,4,2,1 the rest will be another loop of the values 1, 4, and 2.

This equation was formed in 1937 by a man named Lothar Collatz which is why it is referred to as the Collatz Conjecture.

3. The Erdős-Strauss Conjecture

Equation: 4/n=1/a+1/b+1/c

  • a, b and c are positive integers.

This equation aims to see if we can prove that for if n is greater than or equal to 2, then one can write 4*n as a sum of three positive unit fractions.

This equation was formed in 1948 by two men named Paul Erdős and Ernst Strauss which is why it is referred to as the Erdős-Strauss Conjecture.

4. Equation Four

Equation: Use 2(2∧127)-1 – 1 to prove or disprove if it’s a prime number or not?

Looks pretty straight forward, does it? Here is a little context on the problem.

Let’s take a prime number 2. Now, 22 – 1 = 3 which is also a prime number. 25 – 1 = 31 which is also a prime number and so is 27−1=127. 2127 −1=170141183460469231731687303715884105727 is also prime.

5. Goldbach's Conjecture

Equation: Prove that x + y = n

  • where x and y are any two primes

This problem, as relatively simple as it sounds has never been solved. Solving this problem will earn you a free million dollars. This equation was first proposed by Goldbach hence the name Goldbach's Conjecture.

If you are still unsure then pick any even number like 6, it can also be expressed as 1 + 5, which is two primes. The same goes for 10 and 26.

6. Equation Six

Equation: Prove that (K)n = JK1N(q)JO1N(q)

  • Where O = unknot (we are dealing with  knot theory )
  • (K)n  =  Kashaev's invariant of K for any K or knot
  • JK1N(q) of K is equal to N- colored Jones polynomial
  • We also have the volume of conjecture as (EQ3)
  • Here vol(K)  =  hyperbolic volume

This equation tries to portray the relationship between  quantum invariants  of knots and  the hyperbolic geometry  of  knot complements . Although this equation is in mathematics, you have to be a physics familiar to grasp the concept.

7. The Whitehead Conjecture

Equation: G = (S | R)

  • when CW complex K (S | R) is aspherical
  • if π2 (K (S | R)) = 0

What you are doing in this equation is prove the claim made by Mr.  Whitehead  in 1941 in  an algebraic topology  that every subcomplex of an  aspherical   CW complex  that is connected and in two dimensions is also spherical. This was named after the man, Whitehead conjecture.

8. Equation Eight

Equation: (EQ4)

  • Where Γ = a  second countable   locally compact group
  • And the * and r subscript = 0 or 1.

This equation is the definition of  morphism  and is referred to as an assembly map.  Check out the  reduced C*-algebra  for more insight into the concept surrounding this equation.

9. The Euler-Mascheroni Constant

Equation: y=limn→∞(∑m=1n1m−log(n))

Find out if y is rational or irrational in the equation above. To fully understand this problem you need to take another look at rational numbers and their concepts.  The character y is what is known as the Euler-Mascheroni constant and it has a value of 0.5772.

This equation has been calculated up to almost half of a trillion digits and yet no one has been able to tell if it is a rational number or not.

10. Equation Ten

Equation: π + e

Find the sum and determine if it is algebraic or transcendental. To understand this question you need to have an idea of  algebraic real numbers  and how they operate. The number pi or π originated in the 17th century and it is transcendental along with e. but what about their sum? So Far this has never been solved.

As you can see in the equations above, there are several seemingly simple mathematical equations and theories that have never been put to rest. Decades are passing while these problems remain unsolved. If you're looking for a brain teaser, finding the solutions to these problems will give you a run for your money.

See the 26 Comments below.

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10 Hard Math Problems That Even the Smartest People in the World Can’t Crack

Posted: December 24, 2023 | Last updated: December 24, 2023

<p>For all of the recent strides we’ve made in the math world—like a <a href="https://www.popularmechanics.com/science/math/a28943849/unsolvable-math-problem/">supercomputer finally solving the Sum of Three Cubes problem</a> that puzzled mathematicians for 65 years—we’re forever crunching calculations in pursuit of deeper numerical knowledge. Some math problems have been challenging us for centuries, and while <a href="https://www.popularmechanics.com/science/math/a31153757/riddles-brain-teasers-logic-puzzles/">brain-busters</a> like these hard math problems may seem impossible, someone is bound to solve ’em eventually. Well, <em>m</em><em>aybe</em>. </p><p>For now, you can take a crack at the hardest math problems known to man, woman, and machine. For more puzzles and brainteasers, check out <a href="https://www.puzzmo.com/today">Puzzmo</a>. <br><br><strong>✅ More from <em>Popular Mechanics</em></strong><em>:</em></p><ul><li><a href="https://www.popularmechanics.com/science/math/a44450386/math-in-mc-escher-art/">To Create His Geometric Artwork, M.C. Escher Had to Learn Math the Hard Way</a></li><li><a href="https://www.popularmechanics.com/science/math/a42862769/understanding-the-fourier-transform/">Fourier Transforms: The Math That Made Color TV Possible</a></li><li><a href="https://www.popularmechanics.com/science/math/a43023757/game-of-trees-math/">The Game of Trees is a Mad Math Theory That Is Impossible to Prove</a></li></ul>

For all of the recent strides we’ve made in the math world—like a supercomputer finally solving the Sum of Three Cubes problem that puzzled mathematicians for 65 years—we’re forever crunching calculations in pursuit of deeper numerical knowledge. Some math problems have been challenging us for centuries, and while brain-busters like these hard math problems may seem impossible, someone is bound to solve ’em eventually. Well, m aybe .

For now, you can take a crack at the hardest math problems known to man, woman, and machine. For more puzzles and brainteasers, check out Puzzmo . ✅ More from Popular Mechanics :

  • To Create His Geometric Artwork, M.C. Escher Had to Learn Math the Hard Way
  • Fourier Transforms: The Math That Made Color TV Possible
  • The Game of Trees is a Mad Math Theory That Is Impossible to Prove

<p>In September 2019, news broke regarding progress on this 82-year-old question, thanks to prolific mathematician Terence Tao. And while the <a href="https://www.popularmechanics.com/science/math/a29033918/math-riddle-collatz-conjecture/">story</a> of Tao’s breakthrough is promising, the problem isn’t fully solved yet.</p><p>A refresher on the <a href="https://mathworld.wolfram.com/CollatzProblem.html">Collatz Conjecture</a>: It’s all about that function f(n), shown above, which takes even numbers and cuts them in half, while odd numbers get tripled and then added to 1. Take any natural number, apply f, then apply f again and again. You eventually land on 1, for every number we’ve ever checked. The Conjecture is that this is true for all natural numbers (positive integers from 1 through infinity).</p><p><strong>✅ <em>Down the Rabbit Hole: <a href="https://www.popularmechanics.com/space/a39679658/lagrange-points-james-webb-space-telescope/">The Math That Helps the James Webb Space Telescope Sit Steady in Space</a></em></strong></p><p>Tao’s recent work is a near-solution to the Collatz Conjecture in some subtle ways. But he most likely can’t adapt his methods to yield a complete solution to the problem, as Tao subsequently explained. So, we might be working on it for decades longer.</p><p>The Conjecture lives in the math discipline known as <a href="https://mathinsight.org/dynamical_system_idea">Dynamical Systems</a>, or the study of situations that change over time in semi-predictable ways. It looks like a simple, innocuous question, but that’s what makes it special. Why is such a basic question so hard to answer? It serves as a benchmark for our understanding; once we solve it, then we can proceed onto much more complicated matters.</p><p>The study of dynamical systems could become more robust than anyone today could imagine. But we’ll need to solve the Collatz Conjecture for the subject to flourish.</p>

1) The Collatz Conjecture

In September 2019, news broke regarding progress on this 82-year-old question, thanks to prolific mathematician Terence Tao. And while the story of Tao’s breakthrough is promising, the problem isn’t fully solved yet.

A refresher on the Collatz Conjecture : It’s all about that function f(n), shown above, which takes even numbers and cuts them in half, while odd numbers get tripled and then added to 1. Take any natural number, apply f, then apply f again and again. You eventually land on 1, for every number we’ve ever checked. The Conjecture is that this is true for all natural numbers (positive integers from 1 through infinity).

✅ Down the Rabbit Hole: The Math That Helps the James Webb Space Telescope Sit Steady in Space

Tao’s recent work is a near-solution to the Collatz Conjecture in some subtle ways. But he most likely can’t adapt his methods to yield a complete solution to the problem, as Tao subsequently explained. So, we might be working on it for decades longer.

The Conjecture lives in the math discipline known as Dynamical Systems , or the study of situations that change over time in semi-predictable ways. It looks like a simple, innocuous question, but that’s what makes it special. Why is such a basic question so hard to answer? It serves as a benchmark for our understanding; once we solve it, then we can proceed onto much more complicated matters.

The study of dynamical systems could become more robust than anyone today could imagine. But we’ll need to solve the Collatz Conjecture for the subject to flourish.

<p>One of the greatest unsolved mysteries in math is also very easy to write. <a href="https://mathworld.wolfram.com/GoldbachConjecture.html">Goldbach’s Conjecture</a> is, “Every even number (greater than two) is the sum of two primes.” You check this in your head for small numbers: 18 is 13+5, and 42 is 23+19. Computers have checked the Conjecture for numbers up to some magnitude. But we need proof for <em>all</em> natural numbers.</p><p>Goldbach’s Conjecture precipitated from letters in 1742 between German mathematician Christian Goldbach and legendary Swiss mathematician <a href="https://www.popularmechanics.com/science/math/a24383/mathematical-constant-e/">Leonhard Euler</a>, considered one of the greatest in math history. As Euler put it, “I regard [it] as a completely certain theorem, although I cannot prove it.”</p><p>✅ <em><strong>Dive In: <a href="https://www.popularmechanics.com/science/math/a43368981/math-of-color-perception/">The Math Behind Our Current Theory of Human Color Perception Is Wrong</a></strong></em></p><p>Euler may have sensed what makes this problem counterintuitively hard to solve. When you look at larger numbers, they have more ways of being written as sums of primes, not less. Like how 3+5 is the only way to break 8 into two primes, but 42 can broken into 5+37, 11+31, 13+29, and 19+23. So it feels like Goldbach’s Conjecture is an understatement for very large numbers.</p><p>Still, a proof of the conjecture for all numbers eludes mathematicians to this day. It stands as one of the oldest open questions in all of math. </p>

2) Goldbach’s Conjecture

One of the greatest unsolved mysteries in math is also very easy to write. Goldbach’s Conjecture is, “Every even number (greater than two) is the sum of two primes.” You check this in your head for small numbers: 18 is 13+5, and 42 is 23+19. Computers have checked the Conjecture for numbers up to some magnitude. But we need proof for all natural numbers.

Goldbach’s Conjecture precipitated from letters in 1742 between German mathematician Christian Goldbach and legendary Swiss mathematician Leonhard Euler , considered one of the greatest in math history. As Euler put it, “I regard [it] as a completely certain theorem, although I cannot prove it.”

✅ Dive In: The Math Behind Our Current Theory of Human Color Perception Is Wrong

Euler may have sensed what makes this problem counterintuitively hard to solve. When you look at larger numbers, they have more ways of being written as sums of primes, not less. Like how 3+5 is the only way to break 8 into two primes, but 42 can broken into 5+37, 11+31, 13+29, and 19+23. So it feels like Goldbach’s Conjecture is an understatement for very large numbers.

Still, a proof of the conjecture for all numbers eludes mathematicians to this day. It stands as one of the oldest open questions in all of math.

<p>Together with Goldbach’s, the Twin Prime Conjecture is the most famous in Number Theory—or the study of natural numbers and their properties, frequently involving prime numbers. Since you've known these numbers since grade school, stating the conjectures is easy.</p><p>When two primes have a difference of 2, they’re called twin primes. So 11 and 13 are twin primes, as are 599 and 601. Now, it's a Day 1 Number Theory fact that there are infinitely many prime numbers. So, are there infinitely many <em>twin</em> primes? The Twin Prime Conjecture says yes. </p><p>Let’s go a bit deeper. The first in a pair of twin primes is, with one exception, always 1 less than a multiple of 6. And so the second twin prime is always 1 more than a multiple of 6. You can understand why, if you’re ready to follow a bit of heady Number Theory. </p><p>✅ <strong><em>Keep Learning: <a href="https://www.popularmechanics.com/science/math/a35681141/graph-theory-better-computer-chip/">If We Draw Graphs Like This, We Can Change Computers Forever</a></em></strong></p><p>All primes after 2 are odd. Even numbers are always 0, 2, or 4 more than a multiple of 6, while odd numbers are always 1, 3, or 5 more than a multiple of 6. Well, one of those three possibilities for odd numbers causes an issue. If a number is 3 more than a multiple of 6, then it has a <a href="https://www.popularmechanics.com/science/math/a30152083/solve-quadratic-equations/">factor</a> of 3. Having a factor of 3 means a number isn’t prime (with the sole exception of 3 itself). And that's why every third odd number can't be prime. </p><p>How’s your head after that paragraph? Now imagine the headaches of everyone who has tried to solve this problem in the last 170 years. </p><p>The good news is that we’ve made some promising progress in the last decade. Mathematicians have managed to tackle closer and closer versions of the Twin Prime Conjecture. This was their idea: Trouble proving there are infinitely many primes with a difference of 2? How about proving there are infinitely many primes with a difference of 70,000,000? That was cleverly proven in 2013 by Yitang Zhang at the University of New Hampshire.</p><p>For the last six years, mathematicians have been improving that number in Zhang’s proof, from millions down to hundreds. Taking it down all the way to 2 will be the solution to the Twin Prime Conjecture. The <a href="https://www.math.ucla.edu/news/terry-tao-phd-small-and-large-gaps-between-primes">closest we’ve come</a>—given some subtle technical assumptions—is 6. Time will tell if the last step from 6 to 2 is right around the corner, or if that last part will challenge mathematicians for decades longer.</p>

3) The Twin Prime Conjecture

Together with Goldbach’s, the Twin Prime Conjecture is the most famous in Number Theory—or the study of natural numbers and their properties, frequently involving prime numbers. Since you've known these numbers since grade school, stating the conjectures is easy.

When two primes have a difference of 2, they’re called twin primes. So 11 and 13 are twin primes, as are 599 and 601. Now, it's a Day 1 Number Theory fact that there are infinitely many prime numbers. So, are there infinitely many twin primes? The Twin Prime Conjecture says yes.

Let’s go a bit deeper. The first in a pair of twin primes is, with one exception, always 1 less than a multiple of 6. And so the second twin prime is always 1 more than a multiple of 6. You can understand why, if you’re ready to follow a bit of heady Number Theory.

✅ Keep Learning: If We Draw Graphs Like This, We Can Change Computers Forever

All primes after 2 are odd. Even numbers are always 0, 2, or 4 more than a multiple of 6, while odd numbers are always 1, 3, or 5 more than a multiple of 6. Well, one of those three possibilities for odd numbers causes an issue. If a number is 3 more than a multiple of 6, then it has a factor of 3. Having a factor of 3 means a number isn’t prime (with the sole exception of 3 itself). And that's why every third odd number can't be prime.

How’s your head after that paragraph? Now imagine the headaches of everyone who has tried to solve this problem in the last 170 years.

The good news is that we’ve made some promising progress in the last decade. Mathematicians have managed to tackle closer and closer versions of the Twin Prime Conjecture. This was their idea: Trouble proving there are infinitely many primes with a difference of 2? How about proving there are infinitely many primes with a difference of 70,000,000? That was cleverly proven in 2013 by Yitang Zhang at the University of New Hampshire.

For the last six years, mathematicians have been improving that number in Zhang’s proof, from millions down to hundreds. Taking it down all the way to 2 will be the solution to the Twin Prime Conjecture. The closest we’ve come —given some subtle technical assumptions—is 6. Time will tell if the last step from 6 to 2 is right around the corner, or if that last part will challenge mathematicians for decades longer.

<p>Today’s mathematicians would probably agree that the <a href="https://mathworld.wolfram.com/RiemannHypothesis.html">Riemann Hypothesis</a> is the most significant open problem in all of math. It’s one of the seven <a href="https://www.claymath.org/millennium-problems/millennium-prize-problems">Millennium Prize Problems</a>, with $1 million reward for its solution. It has implications deep into various branches of math, but it’s also simple enough that we can explain the basic idea right here.</p><p>There is a function, called the Riemann zeta function, written in the image above.</p><p>For each s, this function gives an infinite sum, which takes some basic calculus to approach for even the simplest values of s. For example, if s=2, then 𝜁(s) is the <a href="https://www.math.cmu.edu/~bwsulliv/basel-problem.pdf">well-known series</a> 1 + 1/4 + 1/9 + 1/16 + …, which strangely adds up to exactly 𝜋²/6. When s is a complex number—one that looks like a+b𝑖, using the imaginary number 𝑖—finding 𝜁(s) gets tricky.</p><p>So tricky, in fact, that it’s become the ultimate math question. Specifically, the Riemann Hypothesis is about when 𝜁(s)=0; the official statement is, “Every nontrivial zero of the Riemann zeta function has real part 1/2.” On the plane of complex numbers, this means the function has a certain behavior along a special vertical line. The hypothesis is that the behavior continues along that line infinitely.</p><p>✅ <strong><em>Stay Curious: <a href="https://www.popularmechanics.com/science/math/a29862279/paint-room-using-math/">How to Paint a Room Using Math</a></em></strong></p><p>The Hypothesis and the zeta function come from German mathematician Bernhard Riemann, who described them in 1859. Riemann developed them while studying prime numbers and their distribution. Our understanding of <a href="https://www.popularmechanics.com/science/math/a36014795/mathematicians-discover-new-kind-of-prime-number/">prime numbers</a> has flourished in the 160 years since, and Riemann would never have imagined the power of supercomputers. But lacking a solution to the Riemann Hypothesis is a major setback.</p><p>If the Riemann Hypothesis were solved tomorrow, it would unlock an avalanche of further progress. It would be huge news throughout the subjects of Number Theory and Analysis. Until then, the Riemann Hypothesis remains one of the largest dams to the river of math research. </p>

4) The Riemann Hypothesis

Today’s mathematicians would probably agree that the Riemann Hypothesis is the most significant open problem in all of math. It’s one of the seven Millennium Prize Problems , with $1 million reward for its solution. It has implications deep into various branches of math, but it’s also simple enough that we can explain the basic idea right here.

There is a function, called the Riemann zeta function, written in the image above.

For each s, this function gives an infinite sum, which takes some basic calculus to approach for even the simplest values of s. For example, if s=2, then 𝜁(s) is the well-known series 1 + 1/4 + 1/9 + 1/16 + …, which strangely adds up to exactly 𝜋²/6. When s is a complex number—one that looks like a+b𝑖, using the imaginary number 𝑖—finding 𝜁(s) gets tricky.

So tricky, in fact, that it’s become the ultimate math question. Specifically, the Riemann Hypothesis is about when 𝜁(s)=0; the official statement is, “Every nontrivial zero of the Riemann zeta function has real part 1/2.” On the plane of complex numbers, this means the function has a certain behavior along a special vertical line. The hypothesis is that the behavior continues along that line infinitely.

✅ Stay Curious: How to Paint a Room Using Math

The Hypothesis and the zeta function come from German mathematician Bernhard Riemann, who described them in 1859. Riemann developed them while studying prime numbers and their distribution. Our understanding of prime numbers has flourished in the 160 years since, and Riemann would never have imagined the power of supercomputers. But lacking a solution to the Riemann Hypothesis is a major setback.

If the Riemann Hypothesis were solved tomorrow, it would unlock an avalanche of further progress. It would be huge news throughout the subjects of Number Theory and Analysis. Until then, the Riemann Hypothesis remains one of the largest dams to the river of math research.

<p>The <a href="https://www.claymath.org/millennium-problems/birch-and-swinnerton-dyer-conjecture">Birch and Swinnerton-Dyer Conjecture</a> is another of the six unsolved Millennium Prize Problems, and it’s the only other one we can remotely describe in plain English. This Conjecture involves the math topic known as Elliptic Curves.</p><p>When we recently <a href="https://www.popularmechanics.com/science/math/g29008356/hard-math-problems/">wrote about the toughest math problems that have been solved</a>, we mentioned one of the greatest achievements in 20th-century math: the solution to Fermat’s Last Theorem. Sir Andrew Wiles solved it using Elliptic Curves. So, you could call this a very powerful new branch of math.</p><p>✅ <strong><em>The Latest: <a href="https://www.popularmechanics.com/science/math/a43670398/mathematicians-discovered-something-mind-blowing-about-the-number-15/">Mathematicians Discovered Something Mind-Blowing About the Number 15</a></em></strong></p><p>In a nutshell, an elliptic curve is a special kind of function. They take the unthreatening-looking form y²=x³+ax+b. It turns out functions like this have certain properties that cast insight into math topics like Algebra and Number Theory.</p><p>British mathematicians Bryan Birch and Peter Swinnerton-Dyer developed their conjecture in the 1960s. Its exact statement is very technical, and has evolved over the years. One of the main stewards of this evolution has been none other than Wiles. To see its current status and complexity, check out <a href="http://www.claymath.org/sites/default/files/birchswin.pdf">this famous update</a> by Wells in 2006.</p>

5) The Birch and Swinnerton-Dyer Conjecture

The Birch and Swinnerton-Dyer Conjecture is another of the six unsolved Millennium Prize Problems, and it’s the only other one we can remotely describe in plain English. This Conjecture involves the math topic known as Elliptic Curves.

When we recently wrote about the toughest math problems that have been solved , we mentioned one of the greatest achievements in 20th-century math: the solution to Fermat’s Last Theorem. Sir Andrew Wiles solved it using Elliptic Curves. So, you could call this a very powerful new branch of math.

✅ The Latest: Mathematicians Discovered Something Mind-Blowing About the Number 15

In a nutshell, an elliptic curve is a special kind of function. They take the unthreatening-looking form y²=x³+ax+b. It turns out functions like this have certain properties that cast insight into math topics like Algebra and Number Theory.

British mathematicians Bryan Birch and Peter Swinnerton-Dyer developed their conjecture in the 1960s. Its exact statement is very technical, and has evolved over the years. One of the main stewards of this evolution has been none other than Wiles. To see its current status and complexity, check out this famous update by Wells in 2006.

<p>A broad category of problems in math are called the <a href="https://en.wikipedia.org/wiki/Sphere_packing">Sphere Packing</a> Problems. They range from pure math to practical applications, generally putting math terminology to the idea of stacking many spheres in a given space, like fruit at the grocery store. Some questions in this study have full solutions, while some simple ones leave us stumped, like the Kissing Number Problem.</p><p>When a bunch of spheres are packed in some region, each sphere has a Kissing Number, which is the number of other spheres it’s touching; if you’re touching 6 neighboring spheres, then your kissing number is 6. Nothing tricky. A packed bunch of spheres will have an average kissing number, which helps mathematically describe the situation. But a basic question about the kissing number stands unanswered.</p><p>✅ <strong><em>Miracles Happen: <a href="https://www.popularmechanics.com/science/math/a43510452/mathematicians-discover-new-ramsey-number-upper-bound/">Mathematicians Finally Make a Breakthrough on the Ramsey Number</a></em></strong></p><p>First, a note on dimensions. Dimensions have a specific meaning in math: they’re independent coordinate axes. The x-axis and y-axis show the two dimensions of a coordinate plane. When a character in a <a href="https://www.popularmechanics.com/culture/movies/g156/the-50-greatest-sci-fi-tv-shows/">sci-fi show</a> says they’re going to a different dimension, that doesn’t make mathematical sense. You can’t go to the x-axis.</p><p>A 1-dimensional thing is a line, and 2-dimensional thing is a plane. For these low numbers, mathematicians have proven the maximum possible kissing number for spheres of that many dimensions. It’s 2 when you’re on a 1-D line—one sphere to your left and the other to your right. There’s proof of an exact number for 3 dimensions, although that took until the 1950s.</p><p>Beyond 3 dimensions, the Kissing Problem is mostly unsolved. Mathematicians have slowly whittled the possibilities to fairly narrow ranges for up to 24 dimensions, with a few exactly known, as you can see <a href="https://en.wikipedia.org/wiki/Kissing_number_problem#Some_known_bounds">on this chart</a>. For larger numbers, or a general form, the problem is wide open. There are several hurdles to a full solution, including computational limitations. So expect incremental progress on this problem for years to come. </p>

6) The Kissing Number Problem

A broad category of problems in math are called the Sphere Packing Problems. They range from pure math to practical applications, generally putting math terminology to the idea of stacking many spheres in a given space, like fruit at the grocery store. Some questions in this study have full solutions, while some simple ones leave us stumped, like the Kissing Number Problem.

When a bunch of spheres are packed in some region, each sphere has a Kissing Number, which is the number of other spheres it’s touching; if you’re touching 6 neighboring spheres, then your kissing number is 6. Nothing tricky. A packed bunch of spheres will have an average kissing number, which helps mathematically describe the situation. But a basic question about the kissing number stands unanswered.

✅ Miracles Happen: Mathematicians Finally Make a Breakthrough on the Ramsey Number

First, a note on dimensions. Dimensions have a specific meaning in math: they’re independent coordinate axes. The x-axis and y-axis show the two dimensions of a coordinate plane. When a character in a sci-fi show says they’re going to a different dimension, that doesn’t make mathematical sense. You can’t go to the x-axis.

A 1-dimensional thing is a line, and 2-dimensional thing is a plane. For these low numbers, mathematicians have proven the maximum possible kissing number for spheres of that many dimensions. It’s 2 when you’re on a 1-D line—one sphere to your left and the other to your right. There’s proof of an exact number for 3 dimensions, although that took until the 1950s.

Beyond 3 dimensions, the Kissing Problem is mostly unsolved. Mathematicians have slowly whittled the possibilities to fairly narrow ranges for up to 24 dimensions, with a few exactly known, as you can see on this chart . For larger numbers, or a general form, the problem is wide open. There are several hurdles to a full solution, including computational limitations. So expect incremental progress on this problem for years to come.

<p>The simplest version of the <a href="http://web.stanford.edu/~cm5/unknotting.pdf">Unknotting Problem</a> has been solved, so there’s already some success with this story. Solving the full version of the problem will be an even bigger triumph.</p><p>You probably haven’t heard of the math subject <a href="https://www.popularmechanics.com/preview/eyJpZCI6IjcxNTQwMTNkLWM5NjQtNGRmNi1hMTJmLTE3ZTgzOTQwMGVhOSIsInR5cGUiOiJjb250ZW50IiwidmVyc2lvbiI6MCwidmVyc2lvbmVkIjpmYWxzZSwidmVyc2lvbl9jcmVhdGVkX2F0IjoiIn0=/">Knot Theory</a>. It’s taught in virtually no high schools, and few colleges. The idea is to try and apply formal math ideas, like proofs, to knots, like … well, what you tie your shoes with.</p><p>For example, you might know how to tie a “square knot” and a “granny knot.” They have the same steps except that one twist is reversed from the square knot to the granny knot. But can you prove that those knots are different? Well, knot theorists can.</p><p>✅ <strong><em>Up Next: <a href="https://www.popularmechanics.com/science/math/a30244043/solve-rubiks-cube/">The Amazing Math Inside the Rubik’s Cube</a></em></strong></p><p>Knot theorists’ holy grail problem was an algorithm to identify if some tangled mess is truly knotted, or if it can be disentangled to nothing. The cool news is that this has been accomplished! Several computer algorithms for this have been written in the last 20 years, and some of them <a href="https://www.youtube.com/watch?v=k9ub2mNyd9M">even animate the process</a>.</p><p>But the Unknotting Problem remains computational. In technical terms, it’s known that the Unknotting Problem is in NP, while we don<strong>’</strong>t know if it’s in P. That roughly means that we know our algorithms are capable of unknotting knots of any complexity, but that as they get more complicated, it starts to take an impossibly long time. For now.</p><p>If someone comes up with an algorithm that can unknot any knot in what’s called polynomial time, that will put the Unknotting Problem fully to rest. On the flip side, someone could prove that isn’t possible, and that the Unknotting Problem’s computational intensity is unavoidably profound. Eventually, we’ll find out. </p>

7) The Unknotting Problem

The simplest version of the Unknotting Problem has been solved, so there’s already some success with this story. Solving the full version of the problem will be an even bigger triumph.

You probably haven’t heard of the math subject Knot Theory . It’s taught in virtually no high schools, and few colleges. The idea is to try and apply formal math ideas, like proofs, to knots, like … well, what you tie your shoes with.

For example, you might know how to tie a “square knot” and a “granny knot.” They have the same steps except that one twist is reversed from the square knot to the granny knot. But can you prove that those knots are different? Well, knot theorists can.

✅ Up Next: The Amazing Math Inside the Rubik’s Cube

Knot theorists’ holy grail problem was an algorithm to identify if some tangled mess is truly knotted, or if it can be disentangled to nothing. The cool news is that this has been accomplished! Several computer algorithms for this have been written in the last 20 years, and some of them even animate the process .

But the Unknotting Problem remains computational. In technical terms, it’s known that the Unknotting Problem is in NP, while we don ’ t know if it’s in P. That roughly means that we know our algorithms are capable of unknotting knots of any complexity, but that as they get more complicated, it starts to take an impossibly long time. For now.

If someone comes up with an algorithm that can unknot any knot in what’s called polynomial time, that will put the Unknotting Problem fully to rest. On the flip side, someone could prove that isn’t possible, and that the Unknotting Problem’s computational intensity is unavoidably profound. Eventually, we’ll find out.

<p>If you’ve never heard of <a href="https://plato.stanford.edu/entries/large-cardinals-determinacy/">Large Cardinals</a>, get ready to learn. In the late 19th century, a German mathematician named <a href="https://www.popularmechanics.com/science/a9006/a-brief-history-of-infinities-15551427/">Georg Cantor</a> figured out that infinity comes in different sizes. Some infinite sets truly have more elements than others in a deep mathematical way, and Cantor proved it.</p><p>There is the first infinite size, <a href="https://www.guinnessworldrecords.com/world-records/smallest-infinite-number#:~:text=The%20smallest%20version%20of%20infinity,the%20power%20of%20aleph%200.">the smallest infinity</a>, which gets denoted ℵ₀. That’s a Hebrew letter aleph; it reads as “aleph-zero.” It’s the size of the set of natural numbers, so that gets written |ℕ|=ℵ₀.</p><p>Next, some common sets are larger than size ℵ₀. The major example Cantor proved is that the set of real numbers is bigger, written |ℝ|>ℵ₀. But the reals aren’t that big; we’re just getting started on the infinite sizes.</p><p>✅ <strong><em>More Mind-Blowing Stuff: <a href="https://www.popularmechanics.com/science/math/a43402074/mathematicians-discover-new-13-sided-shape/">Mathematicians Discovered a New 13-Sided Shape That Can Do Remarkable Things</a></em></strong></p><p>For the really big stuff, mathematicians keep discovering larger and larger sizes, or what we call Large Cardinals. It’s a process of pure math that goes like this: Someone says, “I thought of a definition for a cardinal, and I can prove this cardinal is bigger than all the known cardinals.” Then, if their proof is good, that’s the new largest known cardinal. Until someone else comes up with a larger one.</p><p>Throughout the 20th century, the frontier of known large cardinals was steadily pushed forward. There’s now even a beautiful <a href="http://cantorsattic.info/Cantor%27s_Attic">wiki of known large cardinals</a>, named in honor of Cantor. So, will this ever end? The answer is broadly yes, although it gets very complicated.</p><p>In some senses, the top of the large cardinal hierarchy is in sight. Some theorems have been proven, which impose a sort of ceiling on the possibilities for large cardinals. But many open questions remain, and new cardinals have been nailed down as recently as 2019. It’s very possible we will be discovering more for decades to come. Hopefully we’ll eventually have a comprehensive list of all large cardinals.</p>

8) The Large Cardinal Project

If you’ve never heard of Large Cardinals , get ready to learn. In the late 19th century, a German mathematician named Georg Cantor figured out that infinity comes in different sizes. Some infinite sets truly have more elements than others in a deep mathematical way, and Cantor proved it.

There is the first infinite size, the smallest infinity , which gets denoted ℵ₀. That’s a Hebrew letter aleph; it reads as “aleph-zero.” It’s the size of the set of natural numbers, so that gets written |ℕ|=ℵ₀.

Next, some common sets are larger than size ℵ₀. The major example Cantor proved is that the set of real numbers is bigger, written |ℝ|>ℵ₀. But the reals aren’t that big; we’re just getting started on the infinite sizes.

✅ More Mind-Blowing Stuff: Mathematicians Discovered a New 13-Sided Shape That Can Do Remarkable Things

For the really big stuff, mathematicians keep discovering larger and larger sizes, or what we call Large Cardinals. It’s a process of pure math that goes like this: Someone says, “I thought of a definition for a cardinal, and I can prove this cardinal is bigger than all the known cardinals.” Then, if their proof is good, that’s the new largest known cardinal. Until someone else comes up with a larger one.

Throughout the 20th century, the frontier of known large cardinals was steadily pushed forward. There’s now even a beautiful wiki of known large cardinals , named in honor of Cantor. So, will this ever end? The answer is broadly yes, although it gets very complicated.

In some senses, the top of the large cardinal hierarchy is in sight. Some theorems have been proven, which impose a sort of ceiling on the possibilities for large cardinals. But many open questions remain, and new cardinals have been nailed down as recently as 2019. It’s very possible we will be discovering more for decades to come. Hopefully we’ll eventually have a comprehensive list of all large cardinals.

<p>Given everything we know about two of math’s most famous constants, <a href="https://www.popularmechanics.com/science/math/g26630324/what-is-pi/">𝜋</a> and <a href="https://www.popularmechanics.com/science/math/a24383/mathematical-constant-e/">e</a>, it’s a bit surprising how lost we are when they’re added together.</p><p>This mystery is all about <a href="https://openstax.org/books/college-algebra/pages/1-1-real-numbers-algebra-essentials">algebraic real numbers</a>. The definition: A real number is algebraic if it’s the root of some polynomial with integer coefficients. For example, x²-6 is a polynomial with integer coefficients, since 1 and -6 are integers. The roots of x²-6=0 are x=√6 and x=-√6, so that means √6 and -√6 are algebraic numbers.</p><p>✅ <strong><em>Try It Yourself: <a href="https://www.popularmechanics.com/science/math/a43481696/which-cup-will-fill-first-brain-teaser/">Can You Solve This Viral Brain Teaser From TikTok?</a></em></strong></p><p>All rational numbers, and roots of rational numbers, are algebraic. So it might feel like “most” real numbers are algebraic. Turns out, it’s actually the opposite. The antonym to algebraic is transcendental, and it turns out<em> almost all </em>real numbers are transcendental—for certain mathematical meanings of “almost all.” So who’s <a href="https://www.popularmechanics.com/science/math/a30173098/use-algebra-every-day/">algebraic</a>, and who’s transcendental?</p><p>The real number 𝜋 goes back to ancient math, while the number e has been around since the 17th century. You’ve probably heard of both, and you’d think we know the answer to every basic question to be asked about them, right?</p><p>Well, we do know that both 𝜋 and e are transcendental. But somehow it’s unknown whether 𝜋+e is algebraic or transcendental. Similarly, we don’t know about 𝜋e, 𝜋/e, and other simple combinations of them. So there are incredibly basic questions about numbers we’ve known for millennia that still remain mysterious.</p>

9) What’s the Deal with 𝜋+e?

Given everything we know about two of math’s most famous constants, 𝜋 and e , it’s a bit surprising how lost we are when they’re added together.

This mystery is all about algebraic real numbers . The definition: A real number is algebraic if it’s the root of some polynomial with integer coefficients. For example, x²-6 is a polynomial with integer coefficients, since 1 and -6 are integers. The roots of x²-6=0 are x=√6 and x=-√6, so that means √6 and -√6 are algebraic numbers.

✅ Try It Yourself: Can You Solve This Viral Brain Teaser From TikTok?

All rational numbers, and roots of rational numbers, are algebraic. So it might feel like “most” real numbers are algebraic. Turns out, it’s actually the opposite. The antonym to algebraic is transcendental, and it turns out almost all real numbers are transcendental—for certain mathematical meanings of “almost all.” So who’s algebraic , and who’s transcendental?

The real number 𝜋 goes back to ancient math, while the number e has been around since the 17th century. You’ve probably heard of both, and you’d think we know the answer to every basic question to be asked about them, right?

Well, we do know that both 𝜋 and e are transcendental. But somehow it’s unknown whether 𝜋+e is algebraic or transcendental. Similarly, we don’t know about 𝜋e, 𝜋/e, and other simple combinations of them. So there are incredibly basic questions about numbers we’ve known for millennia that still remain mysterious.

<p>Here’s another problem that’s very easy to write, but hard to solve. All you need to recall is the definition of <a href="https://www.mathsisfun.com/rational-numbers.html">rational numbers.</a></p><p>Rational numbers can be written in the form p/q, where p and q are integers. So, 42 and -11/3 are rational, while 𝜋 and √2 are not. It’s a very basic property, so you’d think we can easily tell when a number is rational or not, right?</p><p>Meet the<a href="http://mathworld.wolfram.com/Euler-MascheroniConstant.html"> Euler-Mascheroni constant</a> 𝛾, which is a lowercase Greek gamma. It’s a real number, approximately 0.5772, with a closed form that’s not terribly ugly; it looks like the image above.</p><p>✅ <strong><em>One More Thing: <a href="https://www.popularmechanics.com/science/math/a43469593/high-schoolers-prove-pythagorean-theorem-using-trigonometry/">Teens Have Proven the Pythagorean Theorem With Trigonometry. That Should Be Impossible</a></em></strong></p><p>The sleek way of putting words to those symbols is “gamma is the limit of the difference of the harmonic series and the natural log.” So, it’s a combination of two very well-understood mathematical objects. It has other neat closed forms, and appears in hundreds of formulas.</p><p>But somehow, we don’t even know if 𝛾 is rational. We’ve <a href="https://www.popularmechanics.com/space/deep-space/a30345230/calculating-interstallar-distances/">calculated</a> it to half a trillion digits, yet nobody can prove if it’s rational or not. The popular prediction is that 𝛾 is irrational. Along with our previous example 𝜋+e, we have another question of a simple property for a well-known number, and we can’t even answer it. </p>

10) Is 𝛾 Rational?

Here’s another problem that’s very easy to write, but hard to solve. All you need to recall is the definition of rational numbers.

Rational numbers can be written in the form p/q, where p and q are integers. So, 42 and -11/3 are rational, while 𝜋 and √2 are not. It’s a very basic property, so you’d think we can easily tell when a number is rational or not, right?

Meet the Euler-Mascheroni constant 𝛾, which is a lowercase Greek gamma. It’s a real number, approximately 0.5772, with a closed form that’s not terribly ugly; it looks like the image above.

✅ One More Thing: Teens Have Proven the Pythagorean Theorem With Trigonometry. That Should Be Impossible

The sleek way of putting words to those symbols is “gamma is the limit of the difference of the harmonic series and the natural log.” So, it’s a combination of two very well-understood mathematical objects. It has other neat closed forms, and appears in hundreds of formulas.

But somehow, we don’t even know if 𝛾 is rational. We’ve calculated it to half a trillion digits, yet nobody can prove if it’s rational or not. The popular prediction is that 𝛾 is irrational. Along with our previous example 𝜋+e, we have another question of a simple property for a well-known number, and we can’t even answer it.

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World's Hardest Maths Problem 'Solved'

Easy as ABC

World's Hardest Maths Problem 'Solved'

If attempting long division at school still gives you nightmares, it might be safer to look away now.

Japanese mathematician Shinichi Mochizuki, of Kyoto University, has published four papers which appear to have a serious claim to be proof of one of the most difficult problems in Mathematics: the abc conjecture.

The problem exists within a strand of pure mathematics known as Number Theory and the conjecture states that where three positive numbers a, b and c, which have no common factor and satisfy a + b = c, exist, then d, the product of the distinct prime factors of abc, is rarely much smaller than c.

See, we said it was easy didn't we?

Mochizuki's work has taken four years, and runs to 500 pages; not only that, but he has created a new mathematical language to explain it, which means that peer-reviewing its accuracy may take some time. However, fellow mathematician Dorian Goldfield, has stated that, if proved, it would amount to "one of the most astounding achievements of mathematics in the 21st Century".

The problem was originally proposed in 1985 and, if Mochizuki has cracked it, it automatically leads to solutions for a great deal of related conditional proofs, including two conjectures related to Fermat (one of which is a conditional proof of his famous Last Theorem - albeit the general proof for this is already established).

Well, this is all well and good, but could he split the bill at a curry for over ten people, including a tip? That's the true test.

[View the papers at Mochizuki's website ]

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A 9-Year-Old Came Up With This Weirdly Tricky Math Problem. Can You Solve It?

This kid is awesome. And maybe evil.

Text, Pink, Font, Handwriting, Textile, Magenta, Art,

  • Puzzles like this require you to look and find the one loose end you can pull to unravel the rest.
  • This kid is awesome.

Novelist Celeste Ng has shared a head-scratching system of equations that her 9-year-old son apparently made up over dinner:

She’s right—using simultaneous equations does get really complicated, and if you’d like to send me your work for the solution, I’d love to see it. But the system is also kind of a trick. Like a sudoku or crossword puzzle, there’s one easier clue that lets you begin to unravel the rest.

Text, Font, Pink, Handwriting, Line, Writing, Heart, Magenta, Paper product, Love,

Can you see it? The secret this time is in the second line.

Text, Pink, Font, Handwriting, Line, Writing, Illustration, Parallel, Smile, Magenta,

Having a squared term is like a golden ticket. We immediately know that we're dealing with one of just a handful of squares that could fit the bill.

Text, Font, Pink, Handwriting, Line, Heart, Love, Pattern, Calligraphy, Paper product,

There could be negative and non-integer solutions to all these things—hey, let me know if you solve them out!—but that didn't seem like something even a sneaky and gifted 9-year-old kid would pull. And if our a value were anything bigger than about 5, the f value would be way out of skew with the others.

I immediately thought " a is 4 and f is 1," because this isn't my first math rodeo and I suspect it's not yours either. The geometry teacher gives you class examples of Pythagorean triples like 3, 4, 5 and 5, 12, 13—not the ugly decimal values you find in real life. When we start to substitute those two values back into the equation system, everything else falls into place.

Text, Font, Pink, Handwriting, Line, Writing, Calligraphy, Parallel, Smile, Paper product,

So f = 1, making c = 5. If a = 4, then e = 7. Our solution is {4, 5, 7, 1}.

Why is this system so hard to solve by traditional algebra? It's because these aren't simple linear equations. As soon as our hero multiplied a × a and then a × c , things got complicated. A squared term indicates a quadratic equation—in this example it's even the special case of the difference of two squares , where the middle term falls away during cross multiplication. And the ac – e one is more like bilinear, where two variables are multiplied together, which isn't quadratic complicated, but is trickier to solve than a straightforward linear equation set.

I'm a big fan of Ng and her burgeoning math demi-villain, whose recent specific interests include Monopoly and, of course, Lego bricks. Last fall, he fixed his mom's tape dispenser.

Honestly, challenging math puzzles are a big step toward " lawful good " for this smart kid. Kudos to his novelist mom, too, for working on math problems at dinner and encouraging his creativity, which will hopefully help to continue to take down the pretend barrier between being creative and being mathematical.

preview for Testing The Marshmallow Crossbow

Caroline Delbert is a writer, avid reader, and contributing editor at Pop Mech. She's also an enthusiast of just about everything. Her favorite topics include nuclear energy, cosmology, math of everyday things, and the philosophy of it all. 

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Second-grader’s impossible math homework question leaves parents stumped: ‘help this mama out’.

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I’ll be the first to admit that math isn’t my strongest subject.

I’d even go as far as saying I’m pretty much innumerate (illiterate but for numbers – not a real word but go with it.) 

With that being said, as a full-grown adult, I thought I would be able to figure out the ‘impossible’ year two maths question that’s doing the rounds on the internet right now. 

But I was wrong. So, so wrong. 

It turns out, I’m not in the minority here. Hundreds of fellow adults were also scratching their heads trying to solve the task, which has been deemed ‘unsolvable’ by many. 

yellow piece of paper with writing on it

“Please help this mama out!”

It all started when a mom shared her child’s homework question to the  r/askmath  Reddit. 

The question starts off by referencing a previous equation that read: “200 + 60 + 9 = 269.”

Seems simple, right?

But here’s where it gets tricky.

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The next question encourages students to “stretch your thinking” by writing another addition equation.

It instructed: “The equation must have a 1-, a 2- and a 3-digit addend and use all of these digits.”

The digits in question included; 6, 6, 2, 2, 8, 8, 0, 0 and 0.

The mom shared a photo of the question next to some working out that’s been scribbled out. She captioned the post: “Second-grade math question that we can’t figure out. The teacher asked for an answer as well that included the numbers. I am so stuck!! This is probably so easy, but after an hour I’m at my wits’ end! Second grade!!!”

She added: “Please help this mama out.”

“This is year two homework?!”

People then rushed to the comments to give the equation a crack and share their thoughts on the question as a whole. 

“Fasten your seatbelts… we’re going from ‘Are you smarter than a fifth grader?’ to ‘Are you smarter than a second grader?'” one user quipped.

A second critiqued the question, saying: “Okay but what the heck is this supposed to teach the kids? Guess and check?”

“Yeah, that is not something I would expect very many second graders to get. But then again, that is the ‘Stretch your thinking’ question for this worksheet,” mentioned a third user.

A fourth then made the point, “The unnecessary hyphens after the one and two kinda makes the question confusing, especially for a child.”

“What the hell are addends?” someone else asked, echoing everyone’s thoughts. 

“I’m assuming your kid was taught what addends were before this homework was given? So they should have been able to at least explain that part considering that, from the comments, most people don’t have a clue what those are,” someone replied. 

So, you’re probably dying to know the answer by now. And, thankfully, a smart cookie in the comments was able to help out. 

Apparently, it was as simple as just recreating a similar formula to the previous answer. 

“800 + 60 + 2 = 862,” wrote the user. 

The OP praised them, saying, “Why did you make it look so easy?! We were about to start WWIII over here and you just whipped it out like the obvious answer that it is. Thank you for saving my tired brain and also my child’s teacher from a very worded email. You win hero of the day!”

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