## 10 Hard Math Problems That Continue to Stump Even the Brightest Minds

Maybe you’ll have better luck. For now, you can take a crack at the hardest math problems known to man, woman, and machine. ✅ More from Popular Mechanics :

• Euler’s Number Is Seriously Everywhere. Here’s What Makes It So Special
• Fourier Transforms: The Math That Made Color TV Possible
• The Game of Trees is a Mad Math Theory That Is Impossible to Prove ## 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.

## 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.

## 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.

## The Riemann Hypothesis ## What is the simplest equation no one can solve?

3x+1 popularly called the Collatz conjecture is the simplest math problem no one can solve. Even though it’s easy for almost anyone to understand it’s also at the same time impossible to solve. It’s was named after Lothar Collatz in 1973. This problem has many origin stories and many names. It’s also known as the Ulam conjecture, Kakutani’s problem, Thwaites conjecture, Hasse’s algorithm, the Syracuse problem, and simply 3N+1.

This is the most dangerous problem in mathematics, one that young mathematicians are warned not to waste their time on. It’s a simple conjecture that not even the world’s best mathematicians have been able to solve. This math problem is both famous and infamous as no respectable mathematician is still trying to find a solution for it. It’s thought to be unsolvable.

The Collatz conjecture is the most famous unsolved problem in all of mathematics. This conjecture asks whether repeating two simple arithmetic operations will at some point transform every positive integer into 1. It uses sequences of integers in which each number (or term) is obtained from the previous number as follows: if the previous number is even, the next number is one half of the previous number. If the previous number is odd, the next number is 3 times the previous number plus 1. The conjecture is that these sequences always reach 1, no matter which positive integer is chosen to start the sequence.

Consider the following operation on an arbitrary positive integer:

If the number is even, divide it by two. If the number is odd, triple it and add one.

Now form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next.

The Collatz conjecture is that this process will eventually reach the number 1, regardless of which positive integer is chosen to start with.

We apply two rules. If the number is odd, we multiply by three and add one. So three times seven is 21, plus one is 22. If the number is even, we divide by two. So 22 divided by two is 11. Now, we keep applying these two rules. 11 is odd, so we multiply by 3, 33, and add 1, 34. Even, divide by two, 17, odd. Multiply by 3, 51, add 1, 52, even. Divide by two, 26, still even. Divide by two, 13, odd. So we multiply by 3, 39, add one, and that’s 40, which is even, so we divide by two, 20, divide by two, 10, divide by two, five, odd. Multiply by three, 15, add one, 16, divide by two that’s eight, and then four, two, and one. Now, one is odd, so we multiply by three and add one, which equals four. But four goes to two, goes to one, so we’re in a loop, and the lowest number is one.

Now, the conjecture is this: every positive integer, if you apply these rules, will eventually end up in the four, two, one loop. 

The two things that are so fascinating about this math problem is that you can start with any number and take it through the iterations and you will get wildly different pathways for each number as some can go very high before getting trapped in the 4-2-1 loop and reverting to 1. The amount of steps that it takes for a number to rise to highs and then revert to 1 also varies widely even with numbers starting next to each other.

When you place the trajectories of the different numbers on charts as they rise and revert back to 1 as they go through the math formula the patterns they create are random. The patterns the numbers make look similar to stocks that revert back to the mean after parabolic run ups in price.

Both are examples of geometric Brownian motion. That means if you take the log and remove the linear trend, the fluctuations are random. It’s like flipping a coin each step. If the coin is heads, the line goes up, tails, it goes down. 3x+1 is just like the random volatility of the stock market. Over long-enough periods, the stock market indexes tend to trend upwards, while 3x+1 eventually trends down to 1.

The below chart shows a histogram of the total stopping times (TST) of the integers from 1 to 100 million where the x-axis represents the stopping time and the y-axis represents the frequency. We can see that the largest stopping time does not exceed 500. The most frequent TST is around 175. This provides strong evidence that the Collatz conjecture is true. Since stopping times seem to always be small, computer programs can be written to determine whether the Collatz sequence of a given integer reaches 1 or not. This method has been used to verify the conjecture for all positive integers up to 5 * 260. If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1. Such a sequence would either enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.

The smallest  i  such that  a i  <  a 0   is called the stopping time of  n . Similarly, the smallest  k  such that  a k  = 1  is called the total stopping time of  n .  If one of the indexes  i  or  k  doesn’t exist, we say that the stopping time or the total stopping time, respectively, is infinite. 

## Why is 3x+1 impossible?

What is the longest collatz sequence.

The longest Collatz chain below five million contains 597 elements and starts with 3732423. 

## Can the Collatz conjecture be solved?

The Collatz conjecture is a brain teasing piece of trivia. The mathematical methods used to determine an answer are very interesting. Mathematicians still don’t really know how to decide the true value of the Collatz conjecture with the methods we know to calculate a solution. Every possible answer will involve either a novel application of our current processes or new methods that might also apply to other math problems. That’s why this problem is so interesting. ## Related Posts

The top 10 mathematics of trading, how to calculate roi.

## Some Math Problems Seem Impossible. That Can Be a Good Thing.

November 18, 2020 BIG MOUTH for Quanta Magazine

## Introduction

Construct a convex octagon with four right angles.

It probably says a lot about me as a teacher that I assign problems like this. I watch as students try to arrange the right angles consecutively. When that doesn’t work, some try alternating the right angles. Failing again, they insert them randomly into the polygon. They scribble, erase and argue. The sound of productive struggle is music to a teacher’s ears.

Then they get suspicious and start asking questions. “You said four right angles. Did you really mean three?” “Are you sure you meant to say convex?” “Four right angles would basically make a rectangle. How can we get four more sides in our octagon?” I listen attentively, nodding along, acknowledging their insights.

Finally someone asks the question they’ve been tiptoeing around, the question I’ve been waiting for: “Wait, is this even possible?”

This question has the power to shift mindsets in math. Those thinking narrowly about specific conditions must now think broadly about how those conditions fit together. Those working inside the system must now take a step back and examine the system itself. It’s a question that’s been asked over and over in the history of math, by those working on problems ranging from squaring the circle to circumambulating the city of Königsberg . And it’s a question that helps us shape what mathematics is and how we understand it.

For example, finding an octagon with certain properties is a very different mathematical task than showing that no such octagon could possibly exist. In playing around with different octagons, we might just stumble upon one that has four right angles.

But luck doesn’t play a role in proving that such an octagon can’t exist. It takes deep knowledge, not just of polygons, but of mathematics itself. To consider impossibility, we need to understand that just asserting that a thing exists doesn’t make it so. Mathematical definitions, properties and theorems all live in a tension born of interconnectedness. In trying to imagine our octagon with four right angles, we work inside those interconnected rules.

But to realize our octagon is impossible, we need to step back and look at the big picture. What mathematical and geometric principles might be violated by an octagon with four right angles? Here, the polygon angle sum theorem is a good place to start.

The sum of the interior angles of an n -sided polygon is given by the formula:

S = ( n – 2) × 180º

This is because every n -sided polygon can be cut into ( n − 2) triangles, each with total internal angles of 180º.

For an octagon, this means the interior angles add up to (8 – 2) × 180º = 6 × 180º = 1080º. Now, if four of those angles are right, each with a measure of 90º, that accounts for 4 × 90º = 360º of the angle sum, which leaves 1080º – 360º = 720º to divide up among the octagon’s remaining four angles.

That means the average measure of those four remaining angles must be:

$latex \frac{720º}{4}$ = 180º

But the interior angles of a convex polygon must each measure less than 180º, so this is impossible. A convex octagon with four right angles cannot exist.

Proving impossibility in this way requires stepping back and seeing how different mathematical rules — like the polygon angle sum formula and the definition of a convex polygon — exist in tension. And since proofs of impossibility rely on thinking broadly across rules, there’s often more than one way to construct the proof.

If an octagon had four right angles, walking around just those angles would bring us full circle: It would be as though we had walked completely around a rectangle. This insight leads us to a rule that gives us a different proof of impossibility. It is known that the sum of the exterior angles of a convex polygon is always 360º. Since an exterior angle of a right angle is also a right angle, our four right angles would take up the entire 360º of the octagon’s exterior-angle measure. This leaves nothing for the remaining four angles, again establishing that our octagon is impossible.

Proving that something is impossible is a powerful act of mathematics. It shifts our perspective from that of rule follower to that of rule enforcer. And to enforce the rules, you must first understand them. You need to know not just how to apply them, but when they don’t apply. And you also need to be on the lookout for situations where rules might conflict with one another. Our octagon exploration exposes the interplay between polygons, convexity, right angles and angle sums. And it highlights how  S = ( n – 2) × 180º isn’t just a formula: It’s one condition in a world of competing conditions.

Proofs of impossibility can help us better understand all areas of math. In school, probability lessons often begin with flipping lots of imaginary coins. I invite students to create an unfair coin — one that is biased toward coming up heads or tails — that has the following property: When the coin is flipped twice, the results of the two flips are more likely to be different than the same. In other words, you’re more likely to get heads and tails than to get heads and heads or tails and tails.

After some tinkering and a little productive frustration, the students arrive at an interesting hypothesis: Different outcomes are never more likely than the same outcome. Some algebra illuminates this and hints at an underlying symmetry.

Let’s say that the coin is biased toward heads. We’ll call the probability of flipping heads $latex \frac{1}{2}$ + k , where 0 < k ≤ $latex \frac{1}{2}$. The fact that k > 0 guarantees that heads is more likely than tails, which has probability  $latex \frac{1}{2}$ – k , since the two probabilities must add up to 1.

If we flip the coin twice, the probability of getting two heads or two tails will be

$latex \left(\frac{1}{2}+k\right)^{2}+\left(\frac{1}{2}-k\right)^{2}$.

Here we’re adding the probability of getting two heads (on the left) with the probability of getting two tails (on the right). Using algebra we can simplify the probability of getting the same result on both flips:

$latex \left(\frac{1}{2}+k\right)^{2}+\left(\frac{1}{2}-k\right)^{2}$ = $latex \frac{1}{4}$ + k + k ² + $latex \frac{1}{4}$ –   k  +  k ² = $latex \frac{1}{2}$ + 2 k².

Since k > 0 , we know that $latex \frac{1}{2}$ + 2 k ² > $latex \frac{1}{2}$, which means it’s more likely than not that the outcomes of the flips will be the same. In fact, we see that even if  k = 0 (when the coin is fair), the probability of the same outcomes is exactly $latex \frac{1}{2}$, making the probability of different flips also $latex \frac{1}{2}$. The same outcome will never be less likely than different outcomes.

As with the octagon problem, we see competing mathematical tensions at work: Altering the likelihood of getting one side of the coin changes the likelihood of getting the other, and this interconnectedness governs what’s possible in terms of the two-flip outcomes. We expose those tensions by trying to do the impossible.

We can expose these tensions in every area of math. Try to find six consecutive integers that sum to 342, and some perseverance will lead to a better understanding of parity. (The fact that consecutive integers alternate between even and odd affects what their sums can be.) The search for a cubic polynomial with integer coefficients that has three non-real roots will teach you about the importance of complex conjugates — pairs of complex numbers whose product and sum are always real. And if you try to inscribe a non-square rhombus in a circle, you’ll walk away having discovered an important property of cyclic quadrilaterals — that the opposite angles in a quadrilateral whose vertices lie on a circle must sum to 180 degrees.

Confronting the impossible invites us to explore the boundaries of our mathematical worlds. The impossible itself is already a kind of generalization, so it’s only natural to keep generalizing: An octagon can’t have four right angles, but what about a decagon? What about a convex polygon with n > 4 sides? These kinds of questions push against the boundaries of our mathematical worlds and deepen our understanding of them.

If we push further, the impossible can even inspire the creation of new mathematical worlds. To prove the impossibility of squaring the circle — a problem that’s at least 2,000 years old — we needed the modern theory of transcendental numbers that cannot be roots of integer polynomials. To solve the bridges of Königsberg problem, Euler turned islands and bridges into vertices and edges, bringing to life the rich fields of graph theory and network theory, with their many applications. Taking the square root of −1 led to an entirely new system of arithmetic. And the logician Kurt Gödel changed the landscape of math forever when he proved that it’s impossible to prove that everything that is true is true.

So the next time you’re stuck on a math problem, ask yourself: “Is this possible?” Struggling with impossibility could give you a better understanding of what actually is possible. You might even create some new math along the way.

1. Find the area of the triangle whose side lengths are 46, 85 and 38.

2. Let f ( x ) = 2 x ³ + bx ² + cx + d . Find integers b , c and d such that  f  $latex \left(\frac{1}{4}\right)$ = 0.

3. Find a perfect square, all of whose digits are in the set {2, 3, 7, 8}.

It’s fun to use something like Heron’s formula to compute the area of this non-triangle. Interesting questions will follow from this!

Click for Answer 2: There are different ways to establish the impossibility of this polynomial. For example, these conditions violate the rational root theorem, which says that any rational roots of a polynomial must be a ratio of a factor of the constant term divided by a factor of the leading coefficient.

Click for Answer 3: A curious fact about perfect squares shows us that this task is impossible. The units digit of a perfect square can only be 0, 1, 4, 5, 6 or 9. This can be shown by simply squaring every possible units digit and observing the possible results. Since no perfect square can end in 2, 3, 7 or 8, no perfect square exists with only those digits.

Get highlights of the most important news delivered to your email inbox

Quanta Magazine moderates comments to facilitate an informed, substantive, civil conversation. Abusive, profane, self-promotional, misleading, incoherent or off-topic comments will be rejected. Moderators are staffed during regular business hours (New York time) and can only accept comments written in English. ## Next article #### IMAGES

1. How to Solve a Wordy Math Problem (with Pictures) 2. A Trip Back To High School: Can You Solve This Math Problem? 3. Try to Solve this Simple Math Problem 4. You Can’t Solve This Math Problem Without A Calculator 5. This Basic Math Problem Went Viral Because Most People Can't Solve It 6. Awesomequotes4u.com: This Basic Math Problem Went Viral Because Most People Can't Solve It #### VIDEO

1. How To Solve 𝐇𝐞𝐥𝐥𝐢𝐬𝐡 Math Problems 😎 #bhannatmaths #shorts #integration

2. I CAN SOLVE ANY PROBLEM WITH MATHS

3. MATH can SOLVE all your PROBLEMS 🤣😂

4. how to solve maths problems quickly

5. When you can’t solve maths #mathshorts

6. Math Problems Kids Can't Solve: Harder and More Complex Than They Ever Thought!

1. A Step-by-Step Guide to Solving Any Math Problem

Mathematics can be a challenging subject for many students. From basic arithmetic to complex calculus, solving math problems requires logical thinking and problem-solving skills. However, with the right approach and a step-by-step guide, yo...

2. How to Break Down and Solve Complex Math Problems in Your Homework

Math homework can often be a challenging task, especially when faced with complex problems that seem daunting at first glance. However, with the right approach and problem-solving techniques, you can break down these problems into manageabl...

3. Thinking Outside the Box: Creative Approaches to Solve Math Problems

Mathematics can often be seen as a daunting subject, full of complex formulas and equations. Many students find themselves struggling to solve math problems and feeling overwhelmed by the challenges they face.

4. List of unsolved problems in mathematics

You can help by adding missing items with reliable sources. Many mathematical problems have been stated but not yet solved. These problems come from many areas

5. 5 Simple Math Problems No One Can Solve

5 Simple Math Problems No One Can Solve · Collatz Conjecture · Moving Sofa Problem · Perfect Cuboid Problem · Inscribed Square Problem · Happy Ending

6. 10 Hard Math Problems That May Never Be Solved

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

7. The simplest math problem no one can solve

The Collatz Conjecture is the simplest math problem no one can solve — it is easy enough for almost anyone to understand but notoriously

8. Unsolved Problems -- from Wolfram MathWorld

"Do you have a comment or news on conjectures in the article math.CO/0409509

9. The Simplest Math Problem No One Can Solve

The Collatz Conjecture is the simplest math problem no one can solve — it is easy enough for almost anyone to understand but notoriously

10. The Simplest Math Problem No One Can Solve (Collatz Conjecture)

The Collatz conjecture is the most famous unsolved problem in all of mathematics. This conjecture asks whether repeating two simple arithmetic

11. 6 Deceptively Simple Maths Problems That No One Can Solve

The Collatz conjecture is one of the most famous unsolved mathematical problems, because it's so simple, you can explain it to a primary-school-

12. What is the simplest math problem that most people can't solve, and

What are some math problems that look hard but are insanely easy? My favorite is the one allegedly solved by John

13. The Simple Math Problem We Still Can't Solve

Lothar Collatz's conjecture 1910_1990 refers to the infinite sequences in which: starting from an integer that is divided by 2 if it is even and

14. Some Math Problems Seem Impossible. That Can Be a Good Thing

Struggling with math problems that can't be solved helps us better understand the ones we can.