Fall 2008 Problem 6: Add up all those products....
Given a natural number n, mathematicians define n! to be the product of all natural numbers from 1 to n. So, for example, 5! = 1 * 2 * 3 * 4 * 5 = 120.
The question is week is: find all the n such that the sum 1! + 2! + .... + n! is a perfect square. (For example, it's not true for 5 —- the sum 1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153 is not a perfect square.)
Solution
The correct answers are 1 and 3. However, I believe some people were thrown by the assumption of a series and did not give 3. For that reason, I gave credit for either answer.
Jeffery James got the first correct answer (for four points). Brent Krise had the best explanation (four points):
The only n that satisfies this is the number n=3. I started out testing cases of small n. When n=3 the sum was 9, which worked. When n=4 the sum was 33. When n=6 the sum was 973. The easy way of finding these sums was to add n! to the previous sum. For example for n=6, the sum is 6! + 153 = 973, For larger and larger n the last digit in the sum is always three. This is because we are adding a number ending in 3 to a multiple of 10. For example with n=6 we have 153 + 720. Since there is no n such that n^2 is a number with the last digit being a 3, the only number that satisfies the conditions is n=3.
Other correct answers (earning three points) came from: Ashley Helsm, Matthew Chumley, Michaela Berg, and Susan Edwards. Correct solutions from non-Centenary students came from Amy Hammond, Chris Evert, Derek Smith, and Don Dinnerville.