Cache and Carry

USC mathematician Solomon W. Golomb offered this problem in the Pi Mu Epsilon Journal, Fall 1971 (page 241):

Ted: I have two numbers x and y, where x + y = z. The sum of the digits of x is 43 and the sum of the digits of y is 68. Can you tell me the sum of the digits of z?

Fred: I need more information. When you added x and y how many times did you have to carry?

Ted: Let’s see. … It was five times.

Fred: Then the sum of the digits of z is 66.

Ted: That’s right! How did you know?

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Solution by Robert C. Gebhardt (from the Fall 1972 issue, page 349): If no carrying were involved, we could get the answer by just adding the two digit sums together, 43 + 68 = 111. But each carry increases the final digit sum by only 1 rather than the 10 that would otherwise result. So each carry reduces our expected final digit sum by 9. In this case there are five carries, so the digit sum of the total is 43 + 68 – (5 × 9) = 66.

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October 15, 2018October 14, 2018 | Puzzles