The Last Detail

A puzzle by R.P. Cross:

Find the last digit in the evaluation of $\sum_{n = 1}^{100}n!$

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It would take a lot of pencils to work this out by hand, but an insight makes it easy: For any n ≥ 5, n! must have the factors 2 and 5 and hence be a multiple of 10, with zero as a final digit. So to solve the puzzle we only have to work out 1! + 2! + 3! + 4! = 33. The last digit is 3. (“A Property of Large Factorial Numbers,” Mathematical Gazette 81:490 [March 1997], 80-81.)