Math Puzzle Answers

by John P. Pratt
last updated 4 Sep 2015

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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. This is an example of answering the question asked and not getting frustrated because you cannot also answer how many other coins you'd have.

  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. A poll taker was fired after reporting that she interviewed 100 people of whom 78 drink coffee, 71 drink tea, 48 drink both. Why is her report impossible?

    Think about how many people would be in the groups that only drink coffee and only drink tea. If 78 drink coffee and 48 drink both, then 78 - 48 = 30 drink only coffee. Similarly, if 71 drink tea and 48 drink both, then 71 - 48 = 23 drink only tea. Those three groups are mutually exclusive, so they are all different people. Thus there must have been a minimum of 30 + 23 + 48 = 101 people who were interviewed. That is more than the number reported, and wouldn't even count those who drink neither coffee nor tea.

  6. Tom and Betty have the same birthday and are both in their twenties. He is four times as old as she was when he was three times as old as she was when he was twice as old as she was. How old are they?

    This is mostly an exercise in listening carefully and translating the words into the numbers. Multiplying ages by three and four and ending up in the twenties means that Tom and Betty are nearly the same age, so we can just try starting with some small numbers. Consider the possibility that once Tom was 6 and Betty was 3. Then Tom would have been 9 when she was 6, so he would have been 3 times as old as when she was when he was twice as old (6) as she was (3). Similarly now he would be 4 times as old as she was then (6) or 24, and she would now be 21. Those are the only numbers which fulfill the requirements and are both in the twenties.

  7. 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?

    This is a problem for which I don't have a clever solution based on an illuminating perspective. I just solved it by brute force by checking possible solutions. I don't have a proof that this is the best (smallest) answer, other than I could not find a better one. I'm not even sure why I put this on my favorite puzzle list, because this answer is only being posted on 3 Sep 2015.

    My answer is that he gave each of the eleven players 12 cents. Here's how they could spent if differently:

    1: 6 @ 2;
    2: 4 @ 2 + 1 @ 4;
    3: 3 @ 2 + 1 @ 6;
    4: 3 @ 2 + 2 @ 3;
    5: 2 @ 2 + 2 @ 4;
    6: 1 @ 2 + 1 @ 4 + 1 @ 6;
    7: 1 @ 2 + 2 @ 3 + 1 @ 4
    8: 4 @ 3;
    9: 2 @ 3 + 1 @ 6;
    10: 3 @ 4;
    11: 2 @ 6;

    Because this solution had exactly 11 answers and I couldn't find one with so many answers using less than 12 cents, it looks correct to me. For those of the younger generation, I'm using the @ sign with its original meaning: 6 @ (priced at) 2 cents each.

  8. 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?

  9. 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!

  10. 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.

  11. 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

    Answer: This problem is one of perception and there may be many ways to do it. Here's my way. Looking at the most significant digits, E must be = 1 in order for there to be no digit in that slot after the subtraction. Moreover, it must be that borrowing was required to subtract T from L in the fifth positions to get E = 1. That means L = 0 (or 1, but it cannot be 1 because E=1) and T = 9.

    Now looking at the least significant side we see the N - E = T, or N - 1 = T. But T is 9 so that means that N = 0. In these digit substitution problems, it cannot be that both N and L are zero. So there is no solution.

  12. A tournament had A and B divisions, each with from 3 to 9 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, provided that his friend were allowed to change divisions. How many contestants ended up in each division?

    To solve this, it is probably easiest to create a table of how many matches are required for a round robim with n contestants. Anyone who has run a round robin tournament may be able to simple see the answer, but let's create that table.

    If there are n contestants, then each plays n-1 others. The number of matches is n(n-1)/2 because when player A plays B it is the same match as B playing A so we need to divide by 2. Here is a table of the number of matches required for each possible number of contestants.

    Number of Contestants3456789
    Number of Games361015212836

    From the table it is seen that the number of games goes up very quickly with the number of contestants, which is why there is not time to play round robin tournaments with many contestants even though that is the best form of tournament because all contestants play each other. Thus, the answer almost certainly has to be that originally the tournament had the opposite extremes of 3 and 9 players. Sure enough, if there had originally been 3 and 9 contestants in the two divisions, then there would have been 3 + 36 = 39 matches scheduled. Adding another contestant to the division of 3 and letting the friend change divisions would result in divisions with 5 and 8 contestants. That would lead to 10+28 = 38 matches. Looking at any other possibilities would not decrease the number of matches.

  13. 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.

  14. A girl spent exactly one dollar (no sales tax) 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 she buy?

    This is an example of what looks like a simple algebra problem, but then looks impossible because it has an infinite number of answers. But then it is must be realized that only one solution of those answers has all three numbers positive integers.

    Let's first do the algebra. Let n represent the number of pieces of nickel candy, t be the number of pieces of two-cent candy and p be the number of ten for a penny candy. (This is an old problem, but I remember candy for that price!) Then:

    n + t + p = 100
    because the number of pieces adds to 100.

    A second equation can be written from the monetary facts given:"

    5n + 2t + p/10 = 100
    where now the numbers represent cents. That is, the worth adds up to 100 cents. (No tax, again an old problem!)

    In algebra it is known that to solve for unique answers n, t, and p, usually a third equation is needed. But in this case the third equation is that all three numbers must be positive integers. There is a special branch of math called number theory just for solving such questions, but in the end even there a lot of trial and error is used to find the answer. So let's just solve it with algebra and then look at possible integer answers.

    To eliminate a variable multiply the second equation by 10 and subtract the first equation:

    49n + 19t = 900

    There may be many ways to proceed from here. My method was to realize that the integer p had to be either 10, 20, 30, 40, 50, 60, 70, 80, 90. Just trying a few of those possibilities leads to the answer: 11 pieces of the nickel candy, 19 of the 2 cent, and 70 of the 10 for a penny.

    In case algebra is not your thing, I asked this question to my non-mathematical mother, after I had spent an hour doing it with number theory congruences. I had wondered what a layman would do with it. Without picking up pencil and paper she thought for less than a minute and told me the answer. I was astounded. I asked her how she did it. She said that was the only answer that she found that would work. After all my algebra and number theory, that's what I had done too!

  15. Broke Bill enters the bank to cash a check. The teller gets confused and where cents is written gives him paper dollars, and where dollars are written, gives him cents. As Bill gathers the money, he unwittingly drops a coin on the floor, losing it. When Bill gets home and counts the money, he discovers he has exactly twice the amount that was written on the check. What was amount on the check?

    Advanced Variation: If he had dropped and lost two coins of different denominations, then what would the amount on the check have been?

    This is another type of the last problem, where one of the equations is that the numbers are positive integers.

    Using algebra, let d be the number of dollars and c the number of cents on the check. If x is the amount of the coin lost, then:

    100 c + d - x = 2 ( 100d + c), which simplifies to

    98c = 199d + x

    That equation is written in cents. The amount of the check is 100d + c. Twice that amount is the same as the amount of the check with numbers reversed minus the lost amount. The coin value x can only be from the set of numbers 1, 5, 10, 25 or 50. The coefficient 199 is slightly more than twice 98, so c is a little more that twice d.

    Simply plugging in possible numbers leads to the following:

    c = 3, d = 1, x = 95
    c = 5, d = 2, x = 92
    c = 7, d = 3, x = 89
    c = 9, d = 4, x = 86
    c = 11, d = 5, x = 83
    c = 57, d = 28, x = 14
    c = 59, d = 29, x = 11
    c = 61, d = 30, x = 8
    c = 63, d = 31, x = 5
    c = 65, d = 32, x = 2

    In particular, c = 2d +1 and x decreases by three each time. So all that is left is to see when x is equal to the value of a coin. The first such coin is when x = 50, and I leave it as an exercise for the student to see why that won't work. Going down the list, the only answer that works is when d = 31, c = 63 and x = 5. Thus the check was for $31.63 and a nickel was lost.

    The advanced option requires only to look down the list of possible values and see that the only answer is d = 29, c = 59, x = 11 = 10 + 1. Thus a dime and penny were lost and check was for $29.59.

  16. 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 the original guess. This problem is famous because Marilyn vos Savant, the "smartest person in the world," answered this question in her column in Parade magazine briefly. pointing out that the odds go up to 2/3 that the prize is behind the door not first chosen. Then several 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 was right anyway and gave a nice, short proof. I'll give a different proof which my mother used.

    When I asked my non-mathematical 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 the prize 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. Opening a door does not change the original odds of 1 in 3. 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.

    [Note of Sat 28 Aug 2015:] I just looked up the original response of Marilyn and all the guff she got from math professors all over the country saying she did it wrong. It turns out she had done it the same way I just explained and that proof was rejected. Then she made a simple table of all possible outcomes and got the same result that switching gives a 2/3 chance, but still they didn't believe her. Then people around the country played the game themselves as Marilyn said (changing the first guess) and proved to themselves that they'd win 2 in 3 times. You can read it all at here. I just read all of the responses that she shared from the professors. It is astounding to me that all of them cited their credentials as an expert to tell her what they thought the answer was. None of them attempted to prove they were correct, appealing instead to their credentials. But if they believe that high credentials are a legitimate reason to believe someone, why did they not accept her credentials as having scored the highest IQ in the world?? One the other hand, Marilyn gave two air-tight proofs which they could not understand.

  17. 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?

    Tom is twice as old as Howard will be in the future, so Tom is older than Howard. Jack won't be as old as Tom until sometime in the future, so Tom is also older than Jack. So the only question left is who is older, Jack or Howard. It seems obvious that if Tom is twice as old as Howard will be in the future that Jack is almost certainly older than Howard, but we need to prove that.

    We can write an equation for what is given. Let T, J, and H be their current ages. The future time spoken of when Jack is Tom's age will occur in T-J years. Thus we are told that

    T = 2(H+ (T-J)), which simplifies to

    T = 2(J-H)

    Because all three ages cannot be negative numbers, we see that J - H must be positive. (If it were zero then T would also be zero, but that cannot be because Tom is the oldest.) If J-H is positive then Jack is older than Howard.

  18. 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 like this problem because to me the answer is so counterintuitive. After all, it was pure wine put into the water and a mixture returned, so it would seem that there is more wine in the water.

    To see why, the easiest way might be to just use large easy numbers. We are not told how big the cup and spoon are, so we should be able to choose any size we want. Suppose there are six ounces in each cup and a 3-oz. spoon is used. Then after the first transfer, there are 3 oz. of wine left in that cup and 6 oz. of water and 3 oz. of wine in the second cup. After stirring the second cup, which is now 2/3 water and 1/3 wine, 1 oz. of water and 2 oz. of wine are put in the first cup. That makes the first cup 4. oz. of wine and 2 oz. of water, so it is 2/3 wine. Thus there is as much wine in the water (2 oz.) as there is water in the wine (2 oz.).

    Here's how to prove that this result works for any size cup and spoon. Let C be the original amount in each cup, and s be the size transferred in the spoon. Then at first there is an amount C in each. After the first transfer there is an amount s of wine in the water glass. After stirring, a fraction C/(C+s) of the spoon transferred back is water. That is an amount sC/(C+s) of water which ends up in the wine glass. The amount of wine left in the water is s - s(s/(C+s)) because the fraction s/(C+s) of that spoon is wine.

    Again intuitively these amounts don't look the same, but which is larger? How much wine is left in the water glass after the transfer? Put all terms of the common denominator C+s. The amount of wine left in the water is ( s(C+s) - s2)/(C+s) = (sC + s2 - s2)/(C+s)= sC/(C+s). Thus, the amount of wine in the water equals the amount of water in the wine.

  19. 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 draw pictures of every possible way to cut the room apart along the edges and have the spider walk in a straight line between the points with the room's surface laid out flat. You'll see that one of the ways of cutting the room makes a right triangle with legs of 6+12+6 = 24 and 1+30+1 = 32. 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. Or you could just remember that a 3-4-5 triangle is a right triangle and notice that the legs 24 and 32 are in the ratio of 3 to 4. Thus, the length of the hypotenuse, which is the shortest distance from the spider to the fly, is 40.

    This answer is very instructive and so counterintuitive. It means the spider actually heads diagonally toward a corner and then spirals around the four long sides and then comes in from the opposite corner to the fly. If you've ever wrapped a present using ribbon parallel to the sides and had it slip be diagonally around the corners and be loose, then you've seen the ribbon solve this very type of problem!

  20. 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 probability 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.)