Math Riddles

If you love solving math riddles and brain teasers like these, you will fit in great as a math major at Lewis.

Math Riddle #1: Degree Cube



The figure below is a cube. How many degrees are in angle ABC?


Solution:
Draw a segment from A to C. The figure ABC is an equilateral triangle, i.e. all 3 sides are equal in length. If the sides are equal in length, the angles are the same number of degrees. Therefore angle ABC is 60 degrees.



Math Riddle #2: Tennis Ball Balance



7 tennis balls are identical in every way, except that one of them weighs slightly less than the other 6. How can you identify the lightweight ball with no more than two separate weighings using a balance scale?

Solution:
Place 3 balls on each side of the balance scale. If the 2 sides balance out, you have found the lightest ball (the one set aside). If the two sides do not balance out, the lightest ball is clearly one of the 3 balls on the less weighted side. Now for your second weighing, place 1 each of your remaining 3 balls on either side of the scale. If the two side balance out, the lightest ball is the one set aside. If not, the lightest ball is obviously the one on the less weighted side.

Math Riddle #3: Prime Number Formula



An interesting formula for generating a good many (but not all) primes is n2 + n + 41. If we plug in consecutive integers 1, 2, 3, etc., at least up to a certain point, we can generate prime numbers. Give an integer for which the formula fails. 1

Solution:
The formula will fail for 40, resulting in 1681, which is the square of 41. This formula will also fail for 41, since 41 can be factored out. But, it will generate primes if integers from 1 to 39 are used in the formula, and many (but not all) integers greater than 41 as well.

1  Paul Hoffman, Archimedes' Revenge, 1995, Ballantine Books, p. 38-39.

Math Riddle #4: Three Prime Numbers = 100



Three prime numbers add up to 100. Of the three numbers, one of them is more than a third but less than a half of another. Find the three numbers. 2

Since the three primes add up to 100, one of them must be even. The only even prime is 2. Therefore the other two primes must add up to 98. By trial and error, these can be determined to be 31 and 67. So the three primes are 2, 31, and 67.

2 L.H. Longley-Cook, Work this One Out, 1960, Fawcett Publications, p. 34.

Math Riddle #5: The Shoe Salesman



A customer in a shoe store bought a pair of shoes that were on sale for $15.00. He gave the salesman a $20.00 bill. Since he did not have change, the salesman went to an adjoining store and asked the lady in charge to give him change. She obligingly gave him a $10.00 bill and two $5.00 bills. The shoe man then returned and gave the customer his shoes and $5.00 change. The customer left.

Up to this point the story is very ordinary, but here is where "the plot thickens."

After the customer left, the lady who gave the salesman the change came into the store and told him that the $20.00 bill was a counterfeit. He looked at the bill, agreed that it was indeed worthless, and immediately repaid her with a good $20.00 bill.

That night, as he was closing the store, the shoe man began thinking about what he had lost in this series of transactions.

What did he lose?

Solution:
The lady next door broke even. She gave $20.00 to the salesman, but he in turn gave her a good $20.00 bill. Therefore, her role in the problem can be eliminated because she and the shoe man merely exchanged $20.00 in cash.

The man who bought the shoes came in with nothing except a worthless $20.00 bill. He walked out of the store with the $15.00 shoes and $5.00 change. So his "gain" was the $15.00 shoes and the $5.00 bill.

What the customer gained was what the shoe man lost. Since the customer gained the shoes plus $5.00, the shoe man in turn lost the shoes plus $5.00.

Math Riddle #6: Smith-Jones-Robinson Classic



Every fact is important. The puzzle is as follows:

On a train, Smith, Jones, and Robinson are the fireman, brakeman, and engineer but not necessarily in that order. Also on the train are three businessmen who have the same names. They will be referred to as Mr. Smith, Mr. Jones, and Mr. Robinson.
  1. Mr. Robinson lives in Detroit.
  2. The brakeman lives exactly halfway between Chicago and Detroit.
  3. Mr. Jones earns exactly $20,000 per year.
  4. The brakeman's nearest neighbor, one of the passengers, earns exactly three times as much as the brakeman.
  5. Smith beats the fireman at billiards.
  6. The passenger whose name is the same as the brakeman's lives in Chicago.
What is the name of the engineer? 3

Solution:

  1. Mr. Robinson lives in Detroit.
  2. The brakeman lives exactly halfway between Chicago and Detroit.
  3. Mr. Jones earns exactly $20,000 per year.
  4. The brakeman's nearest neighbor, one of the passengers, earns exactly three times as much as the brakeman.
    1. Therefore the brakeman's nearest neighbor is not Mr. Jones because Mr. Jones' salary of $20,000 per year is not a multiple of 3.
  5. Smith beats the fireman at billiards.
    1. Therefore the fireman is either Jones or Robinson.
  6. The passenger whose name is the same as the brakeman's lives in Chicago.
    1. Therefore, since Mr. Robinson lives in Detroit and since the brakeman's nearest neighbor (one of the passengers) is not Mr. Jones (from 4a) then Mr. Smith must be the brakeman's nearest neighbor.
    2. Therefore Mr. Jones lives in Chicago.
    3. Therefore the brakeman's name is Jones (from 6).
    4. Therefore the fireman is Robinson. (from 5a)
    5. Therefore the engineer is Smith.
3  Mathematics for Pleasure, Oswald Jacoby, with William Benson. Copy right 1962, Fawcett World Library, p. 84.

Math Riddle #7: Simplify the Expression



Express the number below in simplified form.



Solution:

Call the number x. Since x keeps "reappearing" in the number, we get the following equation:



If we solve the equation and ignore the extraneous (negative) solution, we get

This number is the ratio of the length of the longer side to the length of the shorter side in the "golden rectangle" which was thought by the Greeks to be ideally proportioned.

Math Riddle #8: Subdividing a Square



Given squares with sides of lengths 1, 4, 7, 8, 9, 10, 14, 15 and 18, put them together to form a rectangle. 4

Solution:



4  L.H. Longley-Cook, Work this One Out, 1960, Fawcett Publications, pp. 45.

Math Riddle #9: Subdividing a Cube



Can a cube be divided into a number of smaller cubes, each of a different volume? 5

Solution:

No. Assuming that such a decomposition exists, the bottom square would be divided into unequal squares by the bottom faces of the cubes. Consider the smallest of these squares; it could not be located on a corner or on a side of the large square since in this case the two squares adjacent to it would protrude beyond it. There would be no possible square that would fit the fourth side and whose side would be larger than that of our allegedly smallest one. We conclude that the smallest square must be located in the middle of the bottom face of the original cube. Consider now the cubes whose sides are all larger on the four sides of the small square. They fence off the top face of the cube whose bottom face is our small square. This top face, therefore, has to be covered by a number of cubes of smaller size to cover the top face by squares. Consider the smallest of these, and repeat the argument. We will come to the conclusion that the process cannot end in a finite number of steps because, by iterating the argument, we will get smaller and smaller squares on successively higher levels. This denies the possibility of a finite decomposition.

5  Marc Kac and Stanislaw Ulam, Mathematics and Logic, 1969, Mentor Books, pp. 32-33