Math Puzzle Answers

by John P. Pratt
last updated 31 Mar 2007

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  1. Three men each paid $10 to share a $30 hotel room. Later, the manager felt bad about overcharging them for a $25 dollar room, so he gave the bellboy 5 one-dollar bills to return to them. The bellboy returned one dollar to each of the men, but kept two for himself. So the men paid $27 for the room, plus $2 to the bellboy for a total of $29. Where did the extra dollar go?

    This is sort of a trick question, but it usually takes people a surprisingly long time to get it. The trick is the word "plus". In reality, the $2 should be subtracted from (not added to) the $27 to get $25 paid for the room. There is no missing dollar.

  2. If you had fifty common U.S. coins that added up to exactly a dollar, how many dimes would you have?

    You would have 2 dimes. You might also have 40 pennies and 8 nickels, or you might have 45 pennies, a quarter and 2 nickels.

  3. A race car is going around a track that is a square, one mile on a side. If it goes 60 mph around the first two sides and 30 mph on the the third side, how fast must it go to average 60 mph all the way around?

    No, the answer is not 90 mph, and it is also not 120 mph. The average time is the total distance divided by the total time. 60 mph is 1 mile per minute. To average one mile per minute around a 4 mile track, it must finish the race in four minutes. It spent one minute on the first leg, another on the second, and 2 minutes on the third. Thus it has used up all four of the minutes needed! Thus, no matter how fast it goes, it cannot average 60 mph all the way around.

  4. If a hiker averages 2 miles/hr uphill and 6 miles/hr coming back the same trail, what is his average speed going both ways?

    This is similar to the last problem, in that you cannot simply average speeds and get 4 mph as the correct answer. The answer is easiest to see if you pick any distance at all for the length of the trail, say 6 miles. Going up takes 3 hours and coming down takes one hour. That is 12 miles in 4 hours, so the average is 3 mph.

  5. The coach, who also taught math, said he'd give each of the eleven boys on the football team the same amount of money to buy candy, provided that each spent the money differently on the 2, 3, 4, or 6 cent candy. If he gave each the smallest amount with which that could be done, how much did each receive?


  6. If 6 anteaters can eat 6 ants in 6 minutes, how many anteaters would it take to eat 100 ants in 100 minutes? Another variation is: If a chicken and 3/4 can lay an egg and 3/4 in a day and 3/4, how many chickens does it take to lay a dozen eggs lay in a week? (Jr. High).

    If 6 can eat 6 in 6 minutes, then those same 6 can eat 12 in 12 minutes, 24 in 24 minutes or 100 in 100 minutes. You can think of it as all 6 as working on one ant in one minute and then going to the next ant.
    As for the chickens and eggs, you can come up with a formula, which I did in my last posted solution, but it gives no insight into how to do the problem or what are the principles at work. One method I use to solve problems is to begin with something I know, which is whole numbers of chickens. Suppose 2 chickens could lay 2 eggs in 2 days. Let's try varying some of those numbers to see what happens. If 2 chickens lay 2 eggs in 2 days, then it takes each chicken two days to lay an egg. So 1 chicken would lay 1 egg in 2 days, and 47 chickens would lay 47 eggs in 2 days. So we learn Rule 1: We can multiply the first two numbers by the same number and keep the third the same. Now let's keep the middle number fixed. How long would it take for 1 chicken to lay 2 eggs? Half as many chickens working would take twice as long to produce the same output. So 1 chicken could lay 2 eggs in 4 days, and 4 chickens could produce 2 eggs in 1 day. So we have Rule 2: If we multiply the number of workers by N and keep the output fixed, then we must divide the time by N. Now suppose we keep the number of chickens fixed. How long would it take 2 chickens to produce 4 eggs? It would take 4 days. And 2 chickens could produce 6 eggs in 6 days. So we have Rule 3: We can keep the first number fixed and multiply the last two by the any number and it will be true. Now let's solve the problem using these rules.
    Method 1: Use two of the rules to change the numbers to the desired ones. If 7/4 chickens can lay 7/4 eggs in 7/4 days, how many chickens would it take to lay 12 eggs in 7 days? We can use either Rule 2 or Rule 3 to change the number of days to 7. Using rule 3, we multiply the last two numbers by 4 and get that 7/4 chickens can produce 7 eggs in 7 days. Now use Rule 1 to change the number of eggs while leaving the days unchanged. We multiply the first two numbers by 12/7 (to get a dozen eggs) and get the answer that 3 chickens can lay a dozen eggs in a week. (Exercise for student: Get same result by using Rule 2.)
    Method 2: This problem was submitted by my friend Moray King who prefers the method of always beginning by applying Rule 1 first to find the output of one worker. Multiplying the first two numbers by 4/7 means that 1 chicken can produce 1 egg in 7/4 days. Now we can apply Rule 3 and multiply by 4 to get one week: 1 chicken can produce 4 eggs in 7 days. And now Rule 1 again and multiply by three to get the desired number of eggs: 3 chickens can lay 12 eggs in 7 days. That took an extra step, but it is easier to remember and conceptually very easy. Which of the Rules would have solved the anteater problem in one step?

  7. Four black cows and three brown cows give as much milk in five days as three black cows and five brown cows give in four days. Which color cow gives the most milk?

    I don't see a way to solve this by inspection because the numbers are too close. One way to solve it is by beginning with my Rule 2 (which keeps production the same) from the last problem. Thus, to see how many of the latter mixture would be needed to produce the same milk in 5 days (5/4 the time), we multiply the cows by 4/5. That means that 2.4 black cows and 4 brown cows produce the same milk in 5 days as do four black and three brown. That means that 1 brown cow is producing as much as 1.6 black cows, so the brown is more.
    Now let's solve it with algebra. The rate of production of each cow times the time equals the amount of milk. Let rB be the rate for Black cows and rb be for brown cows. Then (4*rB + 3*rb)* 5 days = (3*rB + 5*rb)* 4 days. Solving this equation for the ratio of rates rB/rb yields 5/8, which is the same answer. So the brown cow produces 8/5 as much!

  8. How many times a day do the hour and minute hands on a clock line up exactly with each other?

    No, it is not 24. Look at a clock. The hands align at 12:00, then 1:06, 2:12, ... 10:54, but that is all. There is no 11:60 because that is the same ast 12:00. So they align 11 times per revolution of the hour hand, or 22 times per day.

  9. Without writing down any trial solutions, prove that the following digit substitution problem has no solution:
               E L E V E N
               - T H R E E
                 E I G H T


  10. A tournament had A and B divisions, each with from 3 to 10 contestants. Each division was a round-robin, in which every contestant plays every other contestant. One contestant was not allowed to register late because there were already so many matches to be played. However, that decision was reversed when he showed that allowing him to enter would actually decrease the total number of matches by 2, provided that his friend were allowed to change divisions. How many contestants ended up in each division?


  11. Two professors met again after several years and one said he was now married with three children. When the other asked their ages, the first said that the product of their ages is 36. The second asked for another clue and the first asked if he could see the number on the house across the street. When the other said yes, the first said that the sum of their ages equaled that number. The second said he still could not determine their ages. Then the first said that his oldest child has red hair. Finally the second knew their ages. What were their ages and how did he know?

    This is my all time favorite problem. To do it, first write down all the real possibilities that the number on that building might have been. Assuming integer ages you get the following which equal 36 when multiplied:

    Ages = 1,1,36 (sum = 38)
    Ages = 1,2,18 (sum = 21)
    Ages = 1,3,12 (sum = 16)
    Ages = 1,4,9 (sum = 14)
    Ages = 1,6,6 (sum = 13)
    Ages = 2,2,9 (sum = 13)
    Ages = 2,3,6 (sum = 11)
    Ages = 3,3,4 (sum = 10)

    The big clue is that the second professor DID NOT KNOW after having been told the sum equalled the number on the house. Why didn't he know? The only reason would be that the number was thirteen, in which case there are two possible answers. For any other number, the answer is unique and the professor would have known after the second clue. So he asked for a third clue. The clue that the oldest had red hair is really just saying that there is an "oldest", meaning that the older two are not twins. Hence, the answer is that the redhead is 9 years old, and the younger two are both two years old.

  12. A boy spent exactly one dollar and bought exactly 100 pieces of candy, including some at $ .05, $.02, and 10 for $.01. How many pieces of each kind of candy did he buy?

    11 pieces of the nickel candy, 19 of the 2 cent, and 70 of the 10/penny.

  13. On a TV quiz show a contestant is shown three closed doors and told that two of them have nothing behind them, but that one has a new car as a prize behind it. The contestant makes her choice of doors where she thinks the prize is. Then one of the other two doors is opened where there is no prize, and the contestant is asked if she would like to change her guess. Do the odds favor changing the guess? Why or why not?

    The odds favor changing your answer. This problem is famous because Marilyn vos Savant, the "smartest person in the world," answered this question in her column in Parade, in one sentence like I just did. Then several famous math professors challenged her, saying that it made no difference, that the chance became 1/2 for each after the other door was opened. Marilyn turned out to be right and gave a nice, short proof. I'll give a different proof which my mother used.

    When I asked my non-mathmatical ninety-year-old Mother this puzzle, I was astounded that she got the right answer in seconds, reasoning as follows. She said, "the chance is 1 in 3 that it is behind the door she chose, and zero chance that it is behind the one that was opened, so there must be a 2 in 3 chance that it is behind the other one." That is exactly right. To see why, pretend that there were 100 doors instead of three. Suppose they kept opening all 98 other doors, so that only the original door chosen was left and one other. Now would you change doors? There is only a 1/100 chance that the first door was right, and a huge chance (99/100) that the other is right.

  14. If Tom is twice as old as Howard will be when Jack is a old as Tom is now, who is the oldest, next oldest and youngest?


  15. A spoonful is removed from a cup of wine and placed in a cup of water and stirred well. Then a spoonful is removed from the resulting solution and put back into the cup of wine. Is there more water in the wine or vice versa?

    There are equal amounts in each. I'll come back and show why some day.

  16. A spider is one foot from the floor in the middle of 12 foot square end wall of a 30-foot long room. A fly is situated in the middle of the other end wall, 1 foot from the ceiling. What is the shortest walking distance from the spider to the fly?

    No the answer is not 42 (going straight up or down and around). The only way I know to do this problem is to drawn pictures of every possible way to cut the room apart and have the spider walk in a straight line between the points. You have to know the Pythagorean formula that sum of the squares of the legs of a right triangle equals the square of the hypotenuse. You'll know when you have the answer because it comes out in round numbers.

  17. A book has 600 pages and an average of 1 typographical error per page. What is the probability of finding n of them on any one page (for n = 0, 1, 2, or any integer). (Hint: use a Poisson distribution.)

    You need higher math for this one, but I put it in because of how nifty the answer is. It turns out that the probability of n errors per page is 1/(en!) where e = 2.71828 and n! means n x (n-1) x (n-2) ... 1.
    For example, the probabily it zero errors per page is 1/e or 1/2.7, which is also the probabily of 1 error per page. The chance of 2 errors per page is half that and of 3 is 1(3x2) or 1/6 of the chance of 1. If you add up all of the possible probabilities of up to all 600 errors being on one page, the all add up to 1 (or very close to one. Technically, if you all them all the way to infinity, they add to one.)