Favorite challenging math puzzles

In February 2014, I asked my subscribers for their favorite math puzzle. This collection is the result—a list of puzzles chosen by teachers and parents! This page lists the more challenging puzzles. The easier ones are listed on this page.

1. The missing $1
   
   This is a famous and truly brain-bending riddle! It really deceives you and makes you feel like something's not right — yet it can feel hard to put your finger on it and find the error in reasoning.
   
   Three people rent a room at $30. They pay $10 each and go up to the room. The owner realized he charged too much and it was only supposed to be $25. He sends the bell hop up with the $5. Each of the people keeps $1 and they give the bellhop $2 as they can't share it. So now each person has paid $9 for the room (total $27) and the bell hop has $2... where is the other $1???
   
   Click here for the answer
   

   The $1.00 isn't "missing" — it is an error to try to add the $27 and the $2. The $2 is the difference of $5 and $3 — it is not the true change in this transaction but is "change of change".
   
   The people paid $30. They were given $5 back as change, but $2 of that change gets returned back to the owner as "change of change". In the end, the people have paid $27 total, and the owner receives $30 − $5 + $2 = $27, so it all matches.
   
   Look at it this way: while the bell hop is going down the stairs, the owner has in his possession $25.00, each of the three people has $1.00 (x 3 = $3.00) and the bell hop has $2.00. $2 + $3 + $25 = $30.
   
   Then when the bell hop returns the $2 to the owner, the owner has $27 and the three people have $3 in total.
   
   Or look at it this way: What they get returned back are the dollars number 26, 27 and 28, and what they return the bell hop are #29 and #30.
   
   Yet another way to think about the answer to this riddle is to just pretend that the bellhop refunded $3 to the people (rather than giving them $5 and receiving $2 back). If the lawyers get $3 back and each takes $1, then they spent exactly $27 dollars. Hide answer
2. As Easy As Pi Puzzle
   
   Cut up the Greek letter Pi below into five pieces as shown. Then re-arrange the five pieces to make a square. Is there more than one way?
   
   
   The solution is here.
   
   Puzzle Author: Stephen Froggatt
3. Fractional exponents
   
   This puzzle was received by Paula Tyroler (my contest participant) from her friend in Prague.
   
   Using only paper and pencil (no calculator or logarithmic tables), determine which of the following expressions has a greater value: 10^1/10 or 3^1/3.
   
   See the answer

   Solution: If we multiply each number by itself, the value will change, but not the relation of the numbers with respect to size. For example if a > b, then a² > b² and aⁿ > bⁿ. In other words, we can raise both sides of an inequality (or equation) to any positive power and the inequality (equality) will be preserved.
   
   Therefore, if 10^1/10 > 3^1/3, then also (10^1/10)³⁰ > (3^1/3)³⁰.
   Now the left side is (10^1/10)³⁰ = 10³ = 1,000.
   The right side is (3^1/3)³⁰ = 3¹⁰. It is easy to see that this will be a lot greater than 1,000 - just start multiplying 3 × 3 × 3 × 3... by using paper and pencil and it won't take too long to reach a number that is more than 1,000. Or, simplify 3¹⁰ as 9⁵ and multiply 9 repeatedly by itself. If you calculate it completely, you will get 3¹⁰ = 59,049.
   
   Therefore, since 3¹⁰ > 10³, then also 3^1/3 > 10^1/10. You can check that using calculator: 3^1/3 ≈ 1.442 and 10^1/10 ≈ 1.258.
   
   Hide answer
4. Weight of coins
   
   This puzzle is a hard variation on easy riddles involving weighing coins (or balls) using an equal-arm balance.
   
   
   You have 12 coins; one weighs slightly less or slightly more than the others. Using an equal-arm balance and only making three weighings, determine which one is different, and whether it is slightly lighter or heavier. The fact that you don't know whether it is lighter or heavier up front makes this problem much more difficult than other similar problems.
   
   This is a very difficult puzzle, and the author doesn't want the solution posted online. You can email the author for solution. Source: Steve Miller's Math Riddles
5. Milk man's containers
   This is an instructive puzzle in at least two ways: notation helps and working from both ends (given situation, desired result) helps. Give the students a day to chew on it. Originally by Sam Lloyd (from the days of home-delivery milkmen, before refrigerators).
   A milkman has two 10-gallon dairy cans of milk (full). One housewife has a three-gallon pail (empty). One housewife has a five-gallon pail (empty). You can say "three-quart" and "five quart" pail. Doesn't matter. Each cook wants two gallons (quarts) of milk. The milkman has no spare containers and no way to mark any containers. He does not want to pour milk away. How does he make the sale?
   
   Solution

   Here's a solution:
   (10, 10, 0, 0)
   (7, 10, 3, 0)
   (7, 5, 3, 5)
   (7, 8, 0, 5)
   (7, 8, 3, 2)
   (9, 8, 3, 0)
   (9, 8, 0, 3)
   (9, 6, 0, 5)
   (9, 6, 3, 2)
   (10,6, 2, 2)
   
   Notice that vector notation helps a lot. Then you can observe that the second-to-last step must look like (n, 6, m, 2).
   
   Hide answer
6. Divisibility
   
   A teacher wrote a large number on the board and asked the students to tell about the divisors of the number one by one.
   
   The 1st student said, "The number is divisible by 2."
   The 2nd student said, "The number is divisible by 3."
   The 3rd student said, "The number is divisible by 4."
   .
   .
   .
   (and so on)
   The 30th student said, "The number is divisible by 31.
   
   The teacher then commented that exactly two students, who spoke consecutively, spoke wrongly.
   
   Which two students spoke wrongly?
   
   Solution

   Clearly, since the two numbers are consecutive, one of them is even and the other is odd. Let's say the two numbers were 5 and 6. This would mean the secret number was not divisible by 5 nor by 6. It would also mean the number could NOT be divisible by 10, 15, 20, 25 or by 12, 18, 24 or 30. Thus the students who said it was divisible by 10, by 15, etc. would also have spoken wrong. So, the two numbers cannot be 5 and 6.
   
   Similarly, the two numbers cannot be, say, 8 and 9, because then also the students who said it was divisible by 16, 24, 18, and 27 would have been in the wrong.
   
   So we can conclude that these two consecutive numbers cannot have multiples that are less than 31. This eliminates a lot of numbers: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, and 15.
   
   Could it be, say 20 and 21? That will not work, because if this secret number IS divisible by 4 and 10, then it is also divisible by 20. Similarly, since we know the number IS divisible by all the whole numbers from 2 through 15, it must also be divisible by 18, 20, 21, 22, 24, 26, 28, and 30.
   
   This leaves the following number pair: 16 & 17.
   
   Or, you can think of it this way. All prime numbers, except 2, are odd. So, one number should be an odd prime and the other should be the highest power of 2 in the range.
   
   Thus the required two numbers are 2⁴ = 16 and 17, and the two students who spoke wrongly are 15th and 16th.
   
   Hide answer
7. Monty Hall Problem
   
   This is a very famous brain teaser in the form of a probability puzzle loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem is a paradox of the veridical type, because the correct result is so counterintuitive it can seem absurd, but is nevertheless demonstrably true.
   
   Wikipedia has some very interesting background details about the problem.
   
   Play Monty Hall online

   Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 3, and the host, who knows what is behind the doors, opens another door, say No. 1, which has a goat. He then says to you, "Do you want to pick door No. 2?"
   
   Is it to your advantage to switch your choice?
   
   Solution

   You should switch. Keep in mind the host always opens a different door from the door chosen by the player and always reveals a goat—because he knows where the car is hidden.
   
   Contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their choice have only a 1/3 chance. One way to see this is to notice that, 2/3 of the time, the initial choice of the player is a door hiding a goat. When that is the case, the host is forced to open the other goat door, and the remaining closed door hides the car. "Switching" only fails to give the car when the player picks the "right" door (the door hiding the car) to begin with. But, of course, that will only happen 1/3 of the time.
   
   Hide answer
8. The lemonkink puzzle
   
   An interesting little brain teaser!
   
   
   I have two glasses the same size. One contains 100 ml of lemonade and the other contains 100 ml of ink. I take a spoonful of lemonade and stir it into the ink, and then take a spoonful of that mixture and stir it back into the lemonade. Which glass now contains least of the contents of the other one?
   
   The solution can be found at Maths Is Fun.
9. Budget for class outing
   
   This problem calls for trying out different combinations, and it was sent to me actually as two different variations. The other variation had an animal fair where the cost of a bull is $10, the cost of a cow is $2.50, and the cost of a calf is $0.50. The person has $100 and wants to buy 100 animals; how will he do that?
   The grade 8 class is going to the planetarium. There is a budget of $100 for the outing and the planetarium can accommodate 100 people. The admission price is as follows:

   Adults $10
   Chaperones: $2.50
   Students: $0.50
   

   Several Senior students are going along to help the teachers. What is the optimal number of teachers, chaperones and students that will be going if all 100 seats are to be filled and all the $100 is to be spent?
   
   Solution

   4 teachers, 6 chaperones, 90 students
   
   Hide answer
10. A car across the desert
    
    
    A car has to carry an important person across the desert. There is no petrol station in the desert and the car has space only for enough petrol to get it half way across the desert. There are also other identical cars that can transfer their petrol into one another. How can we get this important person across the desert?
    
    The solution can be once again found at Maths Is Fun.
11. The liars and truth-tellers — a logic puzzle
    
    Again, a very famous puzzle. The solution is so simple yet seems to elude the common thinker. Even after knowing the answer it can seem difficult to sort through mentally and it illustrates how we tend to be conditioned to think in linear terms rather than approach simple things in a simple manner.
    There's an island in the South Pacific that is divided into 2 territories, the East side and the West side. Although there is nothing about their appearance that would indicated a difference, it is widely known that the natives born on the East side of the island always tell the truth about everything, while the natives born on the West side always lie, with no exceptions. (It's been argued this habit of being untruthful is instilled at a young age as part of the local 'West side' religion, but that's not been verified since every time a West-sider is asked about it they lie).
    
    One afternoon, in the capital city (which happens to be in the exact center of the island with equal land mass in both territories) a missionary from the U.S. was sitting on a bench in the shade relaxing and noticed across the road was a native leaning against a hut. Being a curious sort, the missionary wondered which side of the island this native was originally from. While he was pondering this, another native walking by decided to rest and sat down on the bench beside the missionary. As they were talking, the missionary expressed his curiosity about where the native across the road was born, so the native he was talking to offered to go find out for him. The native got up, walked across the road and the two natives chatted for a minute. Then the native that had crossed the road came back and said to the missionary, "He said he's from the West side of the island."
    
    Question: Not referring to the native standing beside the hut, but rather the one that crossed the road... was he telling the truth or was he lying?
    
    Do not guess. You must explain exactly and logically how you know for certain if he is being truthful or not... and yes, you have all the information you need to know!
    
    Solution

    He lied. How do we know? Think of it in these terms: if you ask a truth teller (East sider) where he is from, he's going to tell the truth and say he's from the "East side". On the other hand, if you ask a liar (West side) where he is from, he is going to lie and say he's from the "East side" as well. So, regardless of where the native by the hut was from, when asked, his answer had to be that he is from the "East side", but the messenger native said, "He said he's from the West side of the island," so he lied.
    
    Hide answer

The Moscow Puzzles: 359 Mathematical Recreations Our whole family enjoys this book! It is divided into 14 chapters, each with different kinds of puzzles, such as matchstick puzzles, dominoes and dice, divisibility, cross sums and magic squares, "Measure Seven Times Before You Cut" (dissection puzzles), "With Algebra and without It" (tricky word problems), and so on. The puzzles and brainteasers range from simple fun riddles to difficult problems, yet none requiring advanced math. They are almost always illustrated and often presented as amusing stories. The book was originally published in Russia in 1956. Martin Gardner has edited the book to make it as easy as possible for an English-reading public to understand, while carefully retaining nearly all the warmth and humor of the original. Price: About $5 used, about $9 new.

See also: Favorite math puzzles for kids — the easier puzzles collected from my contest. Most only require the four basic operations or not even that, so they work well for elementary school children and on up.