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17 Hard Math Problems That’ll Make Your Head Spin

Meghan Jones

These brain-teasing hard math problems are tricky, but they'll give your brain a workout—and they're really rewarding when you figure one out!

rack em math problem question illustration

Time to test your brain!

These hard math problems aren’t straightforward arithmetic. They’ll challenge you to look at the “problems” a different way and test your logic and problem-solving skills while you’re solving. And if math isn’t your strong suit, take heart—most of these hard math problems just use very simple numbers with only basic operations—addition, subtraction, multiplication, and division. What makes them a challenge is often examining the problem to find out the “trick,” or the pattern and the way the numbers relate to one another. Others require a “fill-in-the-blank” technique that requires you to use trial and error and work backward. Time to put your brainpower to work! Maybe brush up on these easy math tricks you’ll wish you’d always known before you get started.

track and field illustration

Track and field

If each of these runners travels the indicated number of spaces in the same amount of time, at which numbered spot will all of the runners be next to one another?

track and field answer illustration

Answer: Space 19

At first, it seems like this will take forever to figure out—but, in actuality, the runners will all be lined up after only six “moves”! Make a chart indicating where each runner will be after each “move.” To figure that out, all you really need to do is add! Just add the number of spaces each runner advances—three, two, one, and five—to the number they’re already at, going back to Space 1 after 30. Sure enough, they’ll all reach Space 19 at the same time. In the mood for something harder? Keep going—or try these difficult brain teasers that will leave you stumped .

numbers on a chalkboard math illustration

Ones and zeros

Below, you’ll see the binary notation for the decimal numbers 1 through 7. Following the pattern, what is the binary notation for the decimal number 12?

ones and zeroes solution illustration

Answer: 1100

If you went to five digits for this one, you did it wrong (and that’s OK; that’s what I did on my first attempt)! What makes this difficult, and different from most hard math problems, is that it’s so unlike how we’re used to looking at numbers, especially in a sequence. Basically, you’re moving through all of the numbers you can make with  just  ones and zeros, starting with the least (0, represented as four 0s) and getting greater. But you’ll always have four digits, so until you have a four-digit number (1000), the digit(s) to the left of the highest place will be a zero. So the question gives you up through 111 (written as 0111 to be four digits), and since it’s representing 7 and you’re trying to reach 12, just go through the next biggest numbers that use only 1s and 0s. So 8 will be 1000, 9 will be 1001, 10 will be 1010, 11 will be 1011, and 12 will be 1100.

stamps illustration

Which number should be on the bottom left stamp? (Hint: It’s a one-digit number.)

face value answer illustration

Many of these hard math problems don’t require doing difficult math as much as they require identifying relationships between groups of numbers. In this case, don’t look at the horizontal rows of stamps but at the vertical pairs of stamps. The number in the ten’s place of the top stamp, divided by the number in the one’s place of the top stamp, equals the number on the bottom stamp (8 divided by 2 is 4 in the leftmost pair of stamps). So, for the last pair, 2 divided by 2 is 1.

energy saver question illustration

Energy saver

A transport company is greatly reducing the amount of oil it uses—but by how much? What number should be below the final barrel? Plus, try out these brain games that boost your brainpower.

energy saver answer illustration

Tricky—this one doesn’t require arithmetic at all! And thankfully, because those are some wacky, large numbers would not be fun to compute with. Instead, look for patterns in the numbers themselves. The following pattern presents itself: Each number (with the exception of the first, which is arbitrary) is the digits of the number before it, minus the first one, reversed. For instance, the first number is 19247. Take away the first digit, 1, and reverse it, and you get 7429—which, sure enough, is the second number. The number before the empty spot only has two digits, so when you remove the first one, 4, you’re left with just 2. No need to “reverse” a number with only one digit, so there’s your answer.

safe cracker question illustration

What’s the digit you need to input in the space with the question mark to complete the code for the safe? Here’s a hint to save you some stress: The “Safe A08” text on the bottom is meaningless, so don’t waste time trying to figure out if that will help you get the answer!

safe code answer illustration

Here’s another thing you need to know to get this answer: Only the groupings of four numbers, two on top and two on the bottom, relate to one another; the entire top and bottom rows are not a sequence, for instance. Think of it as three pairs of two-digit numbers.

So what’s the pattern? Well, the numbers in the ten’s place are being multiplied by two to yield the number in the one’s place of the  other  two-digit number. For instance, in the first pair, you multiply the 2 by two to get the 4 diagonally across from it, and the 3 to get the corresponding 6. And the final pair actually mirrors the first pair; you’re also multiplying a 2 to get a 4!

symbol sums question illustration

Symbol sums

How can you make this equation accurate by using three of these four mathematical symbols: + – × ÷  (add, subtract, multiply, and divide)? NOTE: For this problem, order of operations doesn’t apply; you’d actually be doing each operation from left to right (you don’t have to multiply or divide first, etc).

symbol sums answer illustration

Answer: – ,  + ,  ÷

Here is the correct equation:  21 – 3 + 18 ÷ 6 = 6. Twenty-one minus 3 is 18, then add 18 to that to get 36. Then divide that by 6 to get the correct answer, 6! If you prefer words over numbers, try out these seriously tricky word puzzles .

Rack em question illustration

Rack ’em

Put these billiard balls around the outside into their correct spots, of corresponding colors, so that the sum of every column, row,  and  diagonal is 49. While you’re pondering this, try out this color-identifying test that only 1 percent of people can ace .

rack em answer illustration

Sometimes hard math problems are just about trial and error and eliminating possibilities that don’t work; it’s a little like Sudoku in that way. But there is a  little  bit of strategy to it; for instance, for the leftmost column  and  the bottommost row, you already have a number that ends in 3, so you’ll need the sum of the other three numbers to end in 6. And tackling the bottom-to-top diagonal first will really help; it’s only missing one ball, so you can immediately figure out what that empty spot is. You also know that you can’t put the 25 ball or the 31 ball in any row, column, or diagonal with the 23 ball; they’d immediately make your sum too big.

number music problem illustration

Number music

For this math exercise, see how fast you can solve these not-so-hard math problems—while blasting your favorite music. Oh—and try not to use a calculator!

number music answer illustration

Most of these are just simple math, or even common knowledge (a fourth of 100)!

do the math problem illustration

Do the math

For this one, fill in the numbers 1 through 9, using each only once, to make all of the vertical and horizontal equations correct.

do the math solution illustration

Here’s another one that probably just takes some time and several incorrect guesses.

strike question illustration

Which pins do you have to knock over to get a total of 100 points?

strike solution illustration

Answer: 13, 39, and 48

This one is tricky because it doesn’t tell you how  many  pins you need to knock over, so it could be any combination of the pins. But that’s the only combo that adds up to exactly 100! Test your visual skills by seeing if you can identify these everyday objects by their close-up pictures .

maximize problem illustration

Which three numbers (0 through 9) do you need to replace the question marks with to have the answer be the largest number possible? (No repeat numbers.)

maximize math problem solution illustration

The largest possible solution for this math problem is 44418, and to yield that result, you need to add 128 to 79894, and then subtract 35604. Though this problem might seem mind-boggling at first, the answers really do make sense; you want to get the largest possible sum and then subtract the smallest possible number from it. If you need a break from the numbers, check out these tricky riddles to test your smarts .

Diary Time math problem illustration

What time is the last appointment?

diary time answer illustration

Answer: 19.50

For this answer, you’ll quickly notice a pattern if you add up the digits in each time; ignore the decimal points. The digits in the first “time” add up to 11; the second, 12; the third, 13; and, well, you’ve now got a seriously simple pattern. So you know that the digits in the fifth time must add up to 15. And when you add them…they already do! So the question mark must be replaced with a 0, to keep the sum at 15. Did you know about these math lessons you’ll actually use in real life?

same same math problem illustration

Just by adding the symbols  + – × ÷ √ (add, subtract, multiply, divide, and square root), can you make all of these equations true? The symbols don’t just have to be where the ellipses are; they can be before the first number too (though there won’t be any negative numbers). And there can be more than one symbol in each ellipsis spot. Finally, you’re allowed to group pairs of operations together with parenthesis.

same same answer illustration

OK, yeah, that might have been the hardest of these hard math problems! Square roots can make things extra confusing, and it takes some serious brainpower and ingenuity to figure out where to place those parentheses. Need a break from these hard math problems? Get a chuckle from these clever math jokes .

build it math problem illustration

Which blocks should go in Space A and Space B?

build it answer illustration

Answer: Block 2 in Space A and Block 0 in Space B

This is one of the slightly easier of these hard math problems (and a nice break after that last one, frankly)! You might notice that all of the complete horizontal rows  and  complete vertical columns add up to 8. Then, you know exactly which numbers you need to make the incomplete rows equal 8, too. See how you fare with this elementary school–level math test .

Number blocks math problem illustration

Number blocks

What number should replace the question mark on the bottom right loop?

number blocks answer illustration

And we’re back to a really tricky one! The number on each loop is the number of blocks that could fit in the middle. That one was seriously tricky—just like this “simple” math question for kids that’s stumped the Internet .

Place 100 math problem illustration

Without writing out every single number, can you figure out which letter the number 100 will be underneath?

place 100 answer illustration

Answer: Letter d

The way I figured this out is by thinking about how there are eight columns, so after eight numbers, a new row starts with the letter “a” again. So I thought about multiples of ten (since 100 is one). Because the rows restart after eight numbers, each multiple of ten will be two “letters” after the previous multiple of ten. For instance, in the table, you can already see that 10 is under the letter “b” and 20 is under the letter “d.” So 30 will be under “f,” 40 will be under “h,” 50 will circle back to “b,” and so on.

it's a sign math problem illustration

It’s a sign

What number should go on the bottom right sign?

it's a sign answer illustration

For this one, all you’re doing is putting minus signs between the top and second signs and between the second and third signs, and then an equal sign between the third and bottom signs. So you’re subtracting, and then subtracting again, to get the bottom numbers. So in the first column, you do 41 – 17 – 14 to get 10. (You can also think of it as adding the second and third signs together than then subtracting that from the first sign.) In the final column, you subtract 21 from 37, and then subtract 5 from that, to get the missing number of 11. If you could figure these out, try out some tricky math riddles only the smartest can get right .

Meghan Jones

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20 Grade-School Math Questions So Hard You'll Wonder How You Graduated

Seriously, who can do these?!

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Unless you grew up to be an engineer, a banker, or an accountant, odds are that elementary and middle school math were the bane of your existence. You would study relentlessly for weeks for those silly standardized tests—and yet, come exam day, you'd still somehow have no idea what any of the equations or hard math problems were asking for. Trust us, we get it.

While logic might lead you to believe that your math skills have naturally gotten better as you've aged, the unfortunate reality is that, unless you've been solving algebra and geometry problems on a daily basis, the opposite is more likely the case.

Don't believe us? Then put your number crunching wisdom to the test with these tricky math questions taken straight from grade school tests and homework assignments and see for yourself.

1. Question: What is the number of the parking space covered by the car?

Car Space Brain Teaser {Brain Games}

This tricky math problem went viral a few years back after it appeared on an entrance exam in Hong Kong… for six-year-olds. Supposedly the students had just 20 seconds to solve the problem!

Answer: 87.

Believe it or not, this "math" question actually requires no math whatsoever. If you flip the image upside down, you'll see that what you're dealing with is a simple number sequence.

2. Question: Replace the question mark in the above problem with the appropriate number.

Grade School Math Problem {Tricky Math Questions}

This problem shouldn't be  too difficult to solve if you play a lot of sudoku.

All of the numbers in every row and column add up to 15! (Also, 6 is the only number not represented out of numbers 1 through 9.)

3. Question: Find the equivalent number.

Equivalent Number Math Problem {Tricky Math Questions}

This problem comes straight from a standardized test given in New York in 2014.

a hand writing out 6th grade math equations on a chalkboard {Tricky Math Questions}

You're forgiven if you don't remember exactly how exponents work. In order to solve this problem, you simply need to subtract the exponents (4-2) and solve for 3 2 , which expands into 3 x 3 and equals 9.

4. Question: How many small dogs are signed up to compete in the dog show?

Dog Math Problem {Tricky Math Questions}

Answer: 42.5 dogs.

Group of dogs

In order to figure out how many small dogs are competing, you have to subtract 36 from 49 and then divide that answer, 13, by 2, to get 6.5 dogs, or the number of big dogs competing. But you're not done yet! You then have to add 6.5 to 36 to get the number of small dogs competing, which is 42.5. Of course, it's not actually possible for half a dog to compete in a dog show, but for the sake of this math problem let's assume that it is.

5. Question: Find the area of the red triangle.

Parallelogram Math Problem {Tricky Math Questions}

This question was used in China to identify gifted 5th graders. Supposedly, some of the smart students were able to solve this in less than one minute.

In order to solve this problem, you need to understand how the area of a parallelogram works. If you already know how the area of a parallelogram and the area of a triangle are related, then adding 79 and 10 and subsequently subtracting 72 and 8 to get 9 should make sense—but if you're still confused, then check out this YouTube video for a more in-depth explanation.

6. Question: How tall is the table?

Table Cat Turtle Math Problem {Tricky Math Questions}

YouTuber MindYourDecisions adapted this mind-boggling math question from a similar one found on an elementary school student's homework in China.

Answer: 150 cm.

Table Equation Answer {Tricky Math Questions|

Since one measurement includes the cat's height and subtracts the turtle's and the other does the opposite, you can essentially just act like the two animals aren't there. Therefore, all you have to do is add the two measurements—170 cm and 130 cm—together and divided them by 2 to get the table's height, 150 cm.

7. Question: If the cost of a bat and a baseball combined is $1.10 and the bat costs $1.00 more than the ball, how much is the ball?

Baseball and Bat {Tricky Math Questions}

This problem, mathematically speaking, is very similar to one of the other ones on this list.

Answer: $0.05.

Think back to that problem about the dogs at the dog show and use the same logic to solve this problem. All you have to do is subtract $1.00 from $1.10 and then divide that answer, $0.10 by 2, to get your final answer, $0.05.

8. Question: When is Cheryl's birthday?

Birthday Math Problem {Tricky Math Questions}

If you're having trouble reading that, see here:

"Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.

May 15           May 16           May 19

June 17           June 18

July 14            July 16

August 14       August 15       August 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

Albert: I don't know when Cheryl's birthday is, but I know that Bernard doesn't not know too.

Bernard: At first I don't know when Cheryl's birthday is, but I know now.

Albert: Then I also know when Cheryl's birthday is.

So when is Cheryl's birthday?"

It's unclear why Cheryl couldn't just tell both Albert and Bernard the month and day she was born, but that's irrelevant to solving this problem.

Answer: July 16.

Confused about how one could possibly find any answer to this question? Don't worry, so was most of the world when this question, taken from a Singapore and Asian Schools Math Olympiad competition, went viral a few years ago. Thankfully, though, the  New York Times  explains step-by-step how to get to July 16, and you can read their detailed deduction here.

9. Question: Find the missing letter.

Crazy Algebra Question, hard math problems

This one comes from a  first grader's  homework.

Answer: The missing letter is J.

When you add together the values given for S, B, and G, the sum comes out to 40, and making the missing letter J (which has a value of 14) makes the other diagonal's sum the same.

10. Question: Solve the equation.

Japanese Math Problem, hard math problems

This problem might look easy, but a surprising number of adults are unable to solve it correctly.

Start by solving the division part of the equation. In order to do that, in case you forgot, you have to flip the fraction and switch from division to multiplication, thus getting 3 x 3 = 9. Now you have 9 – 9 + 1, and from there you can simply work from left to right and get your final answer: 1.

11. Question: Where should a line be drawn to make the below equation accurate?

boy struggling Never Say to a Teacher, hard math problems

5 + 5 + 5 + 5 = 555.

Answer: A line should be drawn on a "+" sign.

555 Solution, hard math problems

When you draw a slanted line in the upper left quadrant of a "+," it becomes the number 4 and the equation thusly becomes 5 + 545 + 5 = 555.

12. Question: Solve the unfinished equation.

Math Brain Teaser, hard math problems

Try to figure out what all of the equations have in common.

Answer: 4 = 256.

The formula used in each equation is 4 x  = Y. So, 4 1  = 4, 4 2  = 16, 4 3  = 64, and 4 4  = 256.

13. Question: How many triangles are in the image above?

Triangles Math, hard math problems

When  Best Life  first wrote about this deceiving question, we had to ask a mathematician to explain the answer!

Answer: 18.

Some people get stumped by the triangles hiding inside of the triangles and others forget to include the giant triangle housing all of the others. Either way, very few individuals—even math teachers—have been able to find the correct answer to this problem. And for more questions that will put your former education to the test, check out these  30 Questions You'd Need to Ace to Pass 6th Grade Geography.

14. Question: Add 8.563 and 4.8292.

6th grade math decimal addition, hard math problems

Adding two decimals together is easier than it looks.

Answer: 13.3922.

Don't let the fact that 8.563 has fewer numberrs than 4.8292 trip you up. All you have to do is add a 0 to the end of 8.563 and then add like you normally would.

15. Question: There is a patch of lily pads on a lake. Every day, the patch doubles in size…

Hanging Lake Colorado Enchanting Hideaways in the U.S., hard math problems

… If it takes 48 days for the patch to cover the entire lake, how long would it take for the patch to cover half of the lake?

Answer: 47 days.

Most people automatically assume that half of the lake would be covered in half the time, but this assumption is wrong. Since the patch of pads  doubles in size every day, the lake would be half covered just one day before it was covered entirely.

16. Question: How many feet are in a mile?

Man holding ruler, hard math problems

This elementary school-level problem is a little less problem solving and a little more memorization.

Answer: 5,280.

This was one of the questions featured on the popular show  Are You Smarter Than a 5th Grader?

17. Question: What value of "x" makes the equation below true?

person overthinking, stressed, hard math problems

-15 + (-5x) = 0

Answer: -3.

You'd be forgiven for thinking that the answer was 3. However, since the number alongside x is negative, we need x to be negative as well in order to get to 0. Therefore, x has to be -3.

18. Question: What is 1.92 divided by 3?

decimal division 6th grade math questions, hard math problems

You might need to ask your kids for help on this one.

Answer: 0.64.

In order to solve this seemingly simple problem, you need to remove the decimal from 1.92 and act like it isn't there. Once you've divided 192 by 3 to get 64, you can put the decimal place back where it belongs and get your final answer of 0.64.

19. Question: Solve the math equation above.

Tough Math Equation, hard math problems

Don't forget about PEMDAS!

Using PEMDAS (an acronym laying out the order in which you solve it: "parenthesis, exponents, multiplication, division, addition, subtraction"), you would first solve the addition inside of the parentheses (1 + 2 = 3), and from there finish the equation as it's written from left to right.

20. Question: How many zombies are there?

6th grade math zombies question, hard math problems

Finding the answer to this final question will require using fractions.

Answer: 34.

Since we know that there are two zombies for every three humans and that 2 + 3 = 5, we can divide 85 by 5 to figure out that in total, there are 17 groups of humans and zombies. From there, we can then multiply 17 by 2 and 3 and learn that there are 34 zombies and 51 humans respectively. Not too bad, right?

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These Are the 10 Hardest Math Problems That Remain Unsolved

The smartest people in the world can’t crack them. Maybe you’ll have better luck.

hard problems to solve

For now, you can take a crack at the hardest math problems known to man, woman, and machine. More from Popular Mechanics :

The Collatz Conjecture

hardest math problems

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).

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

hardest math problems

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.”

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

hardest math problems

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.

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

hardest math problems

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.

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.

The Birch and Swinnerton-Dyer Conjecture

hardest math problems

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.

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.

The Kissing Number Problem

hardest math problems

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.

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.

The Unknotting Problem

hardest math problems

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.

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.

The Large Cardinal Project

hardest math problems

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.

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.

What’s the Deal with 𝜋+e?

hardest math problems

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.

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.

Is 𝛾 Rational?

hardest math problems

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.

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.

Headshot of Dave Linkletter

Dave Linkletter is a Ph.D. candidate in Pure Mathematics at the University of Nevada, Las Vegas. His research is in Large Cardinal Set Theory. He also teaches undergrad classes, and enjoys breaking down popular math topics for wide audiences.

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Choose Your Test

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


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.

Image: Sonia Sevilla /Wikimedia

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.


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

Image: Niytx /DeviantArt

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


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) 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.

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


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


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


which can be rewritten


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 $0.167$.


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


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


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:


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

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


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


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


Volume of a Cylinder


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


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


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

Question 13


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


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


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


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!


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.


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.


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


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.

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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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The hardest viral math problems of the year

The internet seems to have a love-hate relationship with math.

People couldn't get enough of math questions this year as they debated the answers in Twitter threads and parenting forums. But they also lamented how much the complicated problems made their brains hurt.

Here are 11 math problems , brainteasers , and SAT questions that went viral this year.

Only 2% of people were able to figure out this riddle in less than 60 seconds.

hard problems to solve

Adam Spencer , a comedian, a mathematician, and the author of " The Number Games ," created a riddle that only 2% of people could solve within one minute , according to Reader's Digest.

He wrote down the numbers one through 19 out of numerical order:

8, 18, 11, 15, 5, 4, 14, 9, 19, 1, 7, 17, 6, 16, ?, ?, ?, ?, ?

To solve the riddle, you need to figure out how he ordered the first 14 numbers and finish the riddle by adding the last five.

The five missing numbers are two, three, 10, 12, and 13. The pattern is that the existing numbers are listed in alphabetical order.

The completed set of numbers looks like this:

8, 18, 11, 15, 5, 4, 14, 9, 19, 1, 7, 17, 6, 16, 10, 13, 3, 12, 2

This Chinese math question for fifth graders stumped adults.

hard problems to solve

A math question for fifth graders in the Chinese district of Shunqing stumped adults around the world .

The problem translates to: "If a ship had 26 sheep and 10 goats onboard, how old is the ship's captain?"

How is the amount of cargo a ship contains supposed to help you figure out how old the captain is?

The internet had a lot to say about this seemingly impossible math question, which became a trending topic on Twitter.

After the test question went viral, the Shunqing Education Department released a statement saying that the question was meant to gauge "critical awareness and an ability to think independently," according to a BBC translation.

This second grade math problem left the internet seriously confused.

hard problems to solve

The question is: "There are 49 dogs signed up to compete in the dog show. There are 36 more small dogs than large dogs signed up to compete. How many small dogs are signed up to compete?"

Popular reasoning was that if there are 49 dogs total, and there are 36 more small dogs than large dogs, you'd subtract 36 from 49. By that measure, there are 13 large dogs and 36 small dogs, meaning the answer is 36. But seeing as that would mean there are 23 more small dogs than large dogs, that isn't right.

The teacher who gave the question later told Popsugar , "The district worded it wrong." But as it stands, the answer is 42.5 : 49 - 36 = 13, 13 / 2 = 6.5, and 36 + 6.5 = 42.5. Apparently, half of a dog competed at the dog show.

People couldn't agree on the answer to a question about buying and selling horses.

hard problems to solve

Mumsnet user PeerieBreeks shared the riddle on the parenting site, where it racked up nearly 500 comments with other users debating the answer .

Here's the question: "A man buys a horse for $60. He sells the horse for $70. He then buys the horse back for $80. And he sells the horse again for $90. In the end, how much money did the man make or lose? Or did he break even?"

Answers in the Mumsnet thread ranged from making $10, $20, and $30 to breaking even. So what's the solution?

What seems to be throwing people off is the fact that the man sells the horse for $70 and then buys it back for $80, making it look like he spent $10 more. But the correct way to solve the problem is to think of the two transactions as separate: -60 + 70 = 10 and -80 + 90 = 10.

The man makes $10 with each sale, therefore he earns a total of $20.

Author Ed Southall shared a math problem on Twitter that some people managed to get right.

hard problems to solve

Ed Southall, author of " Geometry Snacks ," shared a photo of a pink triangle inside a square and challenged people to figure out how much of the square is shaded pink. Some Twitter users gave up immediately, but others rose to the challenge.

According to Business Insider's quant reporter, Andy Kiersz, the key to solving the problem is the height of the pink triangle.

The area of a triangle is 1/2 (base x height). If we assume that the square is a 1 x 1 unit, we can see that the base of the pink triangle is 1, the length of the square. All we need to figure out is the height.

"The key trick is that the little triangle up top is similar to the pink triangle, which means that the little triangle is just a smaller version of the pink triangle," Kiersz said.

"A property of similar triangles is that the ratio of the triangles' heights will be the same as the ratio of their bases," he said. "Since the pink triangle's base is twice the little triangle's base, its height is also twice the little triangle's height. But we know that the little triangle's height plus the pink triangle's height is 1, so that means the pink triangle's height is 2/3. Plug that on in and we get our area = 1/2 x base x height = 1/2 x 1 x 2/3 = 1/3."

Southall confirmed that the answer is indeed 1/3 .

A math problem meant for 8-year-olds went viral after parents couldn't solve it.

hard problems to solve

Mumsnet user lucysmam turned to the internet for help with her daughter's math assignment.

The problem reads as follows:

"On the coast there are three lighthouses.

The first light shines for 3 seconds, then is off for 3 seconds.

The second light shines for 4 seconds, then is off for 4 seconds.

The third light shines for 5 seconds, then is off for 5 seconds.

All three lights have just come on together.

1) When is the first time all three lights will be off at the same time?

2) When is the next time all three lights will come on together at the same moment?"

Thankfully, YouTube math whiz Presh Talwalkar offered an explanation on his channel, MindYourDecisions.

According to Talwalkar, the easiest way to answer the first question about when the lights will all be off is to map out the intervals for each lighthouse and see where their "off" sections overlap. The answer: after five seconds, when the third light has just turned off and the first and second lights are still off.

To determine when all of the lights will come on together, you need to find the smallest common multiple of the intervals when the lights will be on. The answer to that question is that the lighthouses will all come on together at 120 seconds, or two minutes.

For a more detailed explanation, click here .

A tricky math problem created by Gergely Dudás used ice cream cones instead of variables.

hard problems to solve

Artist Gergely Dudás , who is known for his tricky hidden-object puzzles , shared a math problem on his Facebook page that he illustrated with ice cream cones .

The brainteaser consists of four math equations, each of which adds to or multiplies a numerical sum or product. In place of variables like x or y, however, the brainteaser substitutes ice cream cones that are either empty or have scoops of white ice cream, pink ice cream, or both.

To solve the puzzle, you have to figure out what number the empty ice cream cones, white ice cream scoops, and pink ice cream scoops each represent.

The answer is that the empty ice cream cone represents the number three, the white ice cream scoop represents the number two, and the pink ice cream scoop represents the number one.

A Quora thread of difficult SAT math questions included one people called the "meanest test problem ever."

hard problems to solve

In a Quora thread of the most difficult SAT math problems, one question emerged as " the meanest test problem ever ."

"In a class of p students, the average (arithmetic mean) of the test scores is 70.

In another class of n students, the average of the scores for the same test is 92.

When the scores of the two classes are combined, the average of the test scores is 86.

What is the value of p / n ?"

Talwalkar shared a step-by-step solution to the tough problem in a YouTube video.

There are a few ways to solve it, but Talwalkar presented a simple shortcut.

The first class had an average of 70. That's 16 points below the average score of 86. In other words, 86 - 70 = 16. Since there are p students in the class, the difference from the average is 16 p .

The second class had an average of 92. That's 6 points more than the average of 86. In other words, 92 - 86 = 6. There are n students in this class, so the difference from the average is 6 n .

Because these classes average out together — as the problem says, "when the scores of the two classes are combined" — the deficit of points has to be equal to the surplus of points. Therefore, 16 p is equal to 6 n .

Turning that into an equation, we can easily figure out what p / n is:

p/n = 6/16, or 3/8.

A British mathematician says he solved the "single most important open problem" in math after 160 years, but it still needs to be peer reviewed.

hard problems to solve

The Riemann hypothesis was first posited by Bernhard Riemann in 1859. It states that the distribution of prime numbers might follow a pattern described by an equation called the Riemann zeta function. To solve the hypothesis, you need to find a way to predict the occurrence of every prime number, even though primes have historically been regarded as randomly distributed .

The $1 million prize goes to whoever can prove that the equation applies to all prime numbers. British mathematician Sir Michael Atiyah said he solved the 160-year-old problem , but his solution needs to be peer reviewed before he can take home the prize money.

This brainteaser has a missing number.

hard problems to solve

Similar to Sudoku, this puzzle created by Puzzles9  consists of a rectangle with nine squares, each containing a number, except for the bottom-right square.

Read the first two rows of numbers horizontally, each as one number — 289 and 324.

The pattern is that 17 x 17 = 289 and 18 x 18 = 324. So it stands to reason that the bottom row will be 19 x 19 = 361. Therefore, the missing number is one .

This brainteaser features three pentagons, each with a set of five numbers.

hard problems to solve

This brainteaser created by Puzzles9 presents three pentagons, each with a set of five numbers. The middle pentagon is missing a number in the bottom-right corner .

From left to right, let's label the pentagons A, B, and C. The difference between the numbers in pentagons A and B can be found in Pentagon C in the same location across the board. In other words, Pentagon B - Pentagon A = Pentagon C.

So, the missing number is 10.

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Blog / Math / Hardest Math Questions That Are Surprisingly Easy To Solve!

Hardest Math Questions That Are Surprisingly Easy To Solve!

Wonders with math

Have you ever struggled enough with a math problem only to realize that it was too easy to solve? Many students will definitely vouch for it—a math question looked quite simple when they sat with it, but within minutes they understood that they had no idea how to solve it!

Sometimes you come across some of the most difficult math questions that are both challenging and brain teasers. Even if you are well-prepared for your exam, you will encounter difficult math questions. But here’s the thing: They can be solved in a flash, regardless of how ridiculously difficult they appear.

Yes, you heard that right! If you understand basic math skills and concepts, all you need are some math tricks and practice to solve the problems in a jiffy.

Check out These 10 Hardest Math Questions That will Test Your Logic and Problem-solving Skills

Interestingly, these hard math problems can be solved easily with basic math operators. Ready to test your math skills?

Answer: The ball costs 5 cents. How did that happen? If you were thinking of some other answers to this problem, then this explanation might help you. Well, to simplify the question, the difference between $1 and 10 cents is 90 cents, not $1. The only way for the bat to cost a dollar more than the ball while still having the total cost equal to $1.10 is for the baseball bat to cost $1.05 and the ball to cost 5 cents.

Answer: 9 Every PEMDAS problem leads to a huge disruption of opinions, splitting the masses into various answers while everyone believes that they have the correct answer. So, how was this solved and how did they arrive at “9” as the solution? If you remember the order of operations of PEMDAS from grade school, one generally works through the problem by solving parentheses, then the exponents, multiplication, and division, followed by addition and subtraction. Sometimes, some people interpret PEMDAS differently, and that is when the disagreement begins. According to the PEMDAS rule, one should solve anything inside parentheses, then exponents, and then all multiplication and division, starting from the left to the right, as per the operations.

hard problems to solve

Answer: 6 If you enjoy playing Sudoku, then this hard math problem would have been quite a breeze for you! The rows and columns all add up to 15, and that is how the answer turned out to be 6.

Answer: 4 = 256 Have a close look at the equation again, and find out the common thing among the equations. There is a formula that has been used for these equations. 4^x = Y. So, 4^1 = 4, 4^2 = 16, 4^3 = 64, and 4^4 = 256.

Answer: It will take 47 days for the patch to cover half of the lake. If you assumed that half of the lake would be covered in half of the time, then your assumption could be incorrect. Here, in the math problem, it says that the patch “doubles” in size, which means, given any day, the lily pad was half the size the day before. So if the patch covers up the entire lake on the 48th day, it means the lily pad was half the size of the lake on day 47.

Answer: Well, here’s the trick. When it is 9 a.m., you add five hours to it. Voila, you get 2 p.m. as your answer! Getting smarter every minute, aren’t we?

Answer: East. Well, this one might seem tricky, but the answer is quite simple. The students will first turn 90 degrees in a right turn and then 180 degrees in an about-turn. The students will finally turn 90 degrees in a left turn. Therefore, the students are now facing east.

hard problems to solve

Answer: 18 This is another tricky math question where different people have come up with different answers. While some forget to count the hidden triangles, others forget to count the giant triangle. Why don’t you concentrate on the triangle again and find all the triangles?

Equation: 21 _ 3 _ 18 _ 6 = 6

Answer: 21 – 3 + 18 ÷ 6 = 6 This is more of a symbol math problem than a PEMDAS problem. You can keep testing with the math operators until you find the correct answer.

Answer: 20 times If you carefully read the question, the answer is within the question itself. It’s a tricky one, but this is how it goes: John paints on apartment 8, then he paints on apartment 18 (where the 8 digit comes up again) and, similarly, 28th, 38th, 48th, 58th, 68th, 78th, 80th, 81st, 82nd, 83rd, 84th, 85th, 86th, 87th, 88th, 89th, and 98th apartments.

These were some of the hardest math questions, which probably left your mind boggled. You can always keep improving your logical and math skills with these kinds of math problems that might seem difficult but are easier than you had thought. So, keep testing yourself with more tricky math problems and get your logical juices flowing.

To read more about versatile and new online techniques to improve your math concepts and boost your confidence in solving difficult problems, visit Byju’s FutureSchool blog.


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How to Work Through Hard Math Problems

parent of one of our students wrote today about his daughter’s occasional frustration with the difficulty of some of the problems in our courses. She does fantastic work in our courses , and was easily among the very top students in the class she took with me, and yet she still occasionally hits problems that she can’t solve.

Moreover, she has access to an excellent math teacher in her school who sometimes can’t help her get past these problems, either. (This is no slight to him—I have students bring me problems I can’t solve, too!) Her question: “Why does it have to be so hard?”


The Case for Doing Hard Things

We ask hard questions because so many of the problems worth solving in life are hard. If they were easy, someone else would have solved them before you got to them. This is why college classes at top-tier universities have tests on which nearly no one clears 70%, much less gets a perfect score. They’re training future researchers, and the whole point of research is to find and answer questions that have never been solved. You can’t learn how to do that without fighting with problems you can’t solve. If you are consistently getting every problem in a class correct, you shouldn’t be too happy — it means you aren’t learning efficiently enough. You need to find a harder class.

The problem with not being challenged sufficiently goes well beyond not learning math (or whatever) as quickly as you can. I think a lot of what we do at AoPS is preparing students for challenges well outside mathematics. The same sort of strategies that go into solving very difficult math problems can be used to tackle a great many problems. I believe we’re teaching students how to think, how to approach difficult problems, and that math happens to be the best way to do so for many people.

The first step in dealing with difficult problems is to accept and understand their importance. Don’t duck them. They will teach you a lot more than a worksheet full of easy problems. Brilliant “Aha!” moments almost always spring from minds cultivated by long periods of frustration. But without that frustration, those brilliant ideas never arise.

Strategies for Difficult Math Problems — and Beyond

Here are a few strategies for dealing with hard problems, and the frustration that comes with them:

Do something . Yeah, the problem is hard. Yeah, you have no idea what to do to solve it. At some point you have to stop staring and start trying stuff. Most of it won’t work. Accept that a lot of your effort will appear to have been wasted. But there’s a chance that one of your stabs will hit something, and even if it doesn’t, the effort may prepare your mind for the winning idea when the time comes.

We started developing an elementary school curriculum months and months before we had the idea that became Beast Academy . Our lead curriculum developer wrote 100–200 pages of content, dreaming up lots of different styles and approaches we might use. Not a one of those pages will be in the final work, but they spurred a great many ideas for content we will use. Perhaps more importantly, it prepared us so that when we finally hit upon the Beast Academy idea, we were confident enough to pursue it.

Simplify the problem . Try smaller numbers and special cases. Remove restrictions. Or add restrictions. Set your sights a little lower, then raise them once you tackle the simpler problem.

Reflect on successes . You’ve solved lots of problems. Some of them were even hard problems! How did you do it? Start with problems that are similar to the one you face, but also think about others that have nothing to do with your current problem. Think about the strategies you used to solve those problems, and you might just stumble on the solution.

A few months ago, I was playing around with some Project Euler problems, and I came upon a problem that (eventually) boiled down to generating integer solutions to c ² = a ² + b ² + ab in an efficient manner. Number theory is not my strength, but my path to the solution was to recall first the method for generating Pythagorean triples. Then, I thought about how to generate that method, and the path to the solution became clear. (I’m guessing some of our more mathematically advanced readers have so internalized the solution process for this type of Diophantine equation that you don’t have to travel with Pythagoras to get there!)

Focus on what you haven’t used yet . Many problems (particularly geometry problems) have a lot of moving parts. Look back at the problem, and the discoveries you have made so far and ask yourself: “What haven’t I used yet in any constructive way?” The answer to that question is often the key to your next step.

Work backwards . This is particularly useful when trying to discover proofs. Instead of starting from what you know and working towards what you want, start from what you want, and ask yourself what you need to get there.

Ask for help . This is hard for many outstanding students. You’re so used to getting everything right, to being the one everyone else asks, that it’s hard to admit you need help. When I first got to the Mathematical Olympiad Program (MOP) my sophomore year, I was in way over my head. I understood very little of anything that happened in class. I asked for help from the professor once — it was very hard to get up the courage to do so. I didn’t understand anything he told me during the 15 minutes he worked privately with me. I just couldn’t admit it and ask for more help, so I stopped asking. I could have learned much, much more had I just been more willing to admit to people that I just didn’t understand. (This is part of why our classes now have a feature that allows students to ask questions anonymously.) Get over it. You will get stuck. You will need help. And if you ask for it, you’ll get much farther than if you don’t.

Start early . This doesn’t help much with timed tests, but with the longer-range assignments that are parts of college and of life, it’s essential. Don’t wait until the last minute — hard problems are hard enough without having to deal with time pressure. Moreover, complex ideas take a long time to understand fully. The people you know who seem wicked smart, and who seem to come up with ideas much faster than you possibly could, are often people who have simply thought about the issues for much longer than you have. I used this strategy throughout college to great success — in the first few weeks of each semester, I worked far ahead in all of my classes. Therefore, by the end of the semester, I had been thinking about the key ideas for a lot longer than most of my classmates, making the exams and such at the end of the course a lot easier.

Take a break . Get away from the problem for a bit. When you come back to it, you may find that you haven’t entirely gotten away from the problem at all — the background processes of your brain have continued plugging away, and you’ll find yourself a lot closer to the solution. Of course, it’s a lot easier to take a break if you start early.

Start over . Put all your earlier work aside, get a fresh sheet of paper, and try to start from scratch. Your other work will still be there if you want to draw from it later, and it may have prepared you to take advantage of insights you make in your second go-round.

Give up . You won’t solve them all. At some point, it’s time to cut your losses and move on. This is especially true when you’re in training, and trying to learn new things. A single difficult problem is usually going to teach you more in the first hour or two than it will in the next six, and there are a lot more problems to learn from. So, set yourself a time limit, and if you’re still hopelessly stuck at the end of it, then read the solutions and move on.

Be introspective . If you do give up and read the solution, then read it actively, not passively. As you read it, think about what clues in the problem could have led you to this solution. Think about what you did wrong in your investigation. If there are math facts in the solution that you don’t understand, then go investigate. I was completely befuddled the first time I saw a bunch of stuff about “mod”s in an olympiad solution — we didn’t have the internet then, so I couldn’t easily find out how straightforward modular arithmetic is! You have the internet now, so you have no excuse. If you did solve the problem, don’t just pat yourself on the back. Think about the key steps you made, and what the signs were to try them. Think about the blind alleys you explored en route to the solution, and how you could have avoided them. Those lessons will serve you well later.

Come back . If you gave up and looked at the solutions, then come back and try the problem again a few weeks later. If you don’t have any solutions to look at, keep the problem alive. Store it away on paper or in your mind.

Richard Feynman once wrote that he would keep four or five problems active in the back of his mind. Whenever he heard a new strategy or technique, he would quickly run through his problems and see if he could use it to solve one of his problems. He credits this practice for some of the anecdotes that gave Feynman such a reputation for being a genius. It’s further evidence that being a genius is an awful lot about effort, preparation, and being comfortable with challenges.

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