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Question: The warden of a circular jail is extremely hyper one day so he begins running around opening cells. The jail has 100 cells numbered from 1 to 100. He runs in a circle and opens all of the cells. Next he runs around and closes every 2nd cell (starting with cell 2, 4, 6 , etc.). If a cell is open he closes it and if a cell is closed he opens it. When he finishes running by all 100 cells he opens/closes every 3rd cell (starting with 3, 6, 9, 12 etc.), then every 4th cell (starting with 4, 8, 12, etc.) and so on. He does this until he goes around and only changes the 100th cell.

When he is done what cells will be open?

Answer: The cells 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 will be open. Only the cells that have two of the same factors (4 = 2 * 2) will be open because they are opened and closed the same number of times (except 1 which is just opened). Since they have the same factor twice their number of factors is odd.

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Question: A man was born on January 1st, 23 B.C. and died January 2nd, 23 A.D. How old did he live to be?

Answer: 45 years old.

There is no year 0 so you can add 23 to 23 but you must subtract one to take year 0 out of consideration: 23 + 23 - 1 = 45 years old. In some cultures people are born 1 years old, in this case they would be 46 years old when they die.

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Question: You have a flashlight that takes 2 working batteries. You have 8 batteries but only 4 of them work.

What is the fewest number of pairs you need to test to guarantee you can get the flashlight on?

Answer: 7. If you break the batteries into 3 groups: Two groups of 3 and one group of 2. By doing this you guarantee that one of the groups has 2 working batteries. Both of the groups of 3 have 3 possible combinations of 2 batteries and the group of 2 only has 1 combination. So, 3 + 3 + 1 = 7 tries at most to find two working batteries.

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Question: How can you make the following true by adding a single line: .9 = 1.

Note: ≠ is not the answer.

Answer: .9 = 1.

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Question: You have an 8 by 8 chess board that has two corner squares diagonal from each other removed. Is there any way to place 31 dominoes of size 2x1 to cover the remaining 62 spaces?

Answer: No. When you remove the two diagonal spaces from the chess board they are both black or both white, making it so that there are two more of the other color. Each domino would cover 1 black and 1 white space. So 31 of them would have to cover 31 white spaces and 31 black spaces but with the corners removed there are 30 of one color and 32 of the other.

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