HMMT 二月 2001 · 代数 · 第 6 题

HMMT February 2001 — Algebra — Problem 6

解析

What is the last digit of 1 + 2 + 3 + · · · + 100 ? Solution: Let L ( d, n ) be the last digit of a number ending in d to the n th power. For n ≥ 1, we know that L (0 , n ) = 0, L (1 , n ) = 1, L (5 , n ) = 5, L (6 , n ) = 6. All numbers ending in odd digits in this series are raised to odd powers; for odd n , L (3 , n ) = 3 or 7, L (7 , n ) = 3 or 7, L (9 , n ) = 9. All numbers ending in even digits are raised to even powers; for even n , L (2 , n ) = 4 or 6, L (4 , n ) = L (6 , n ) = 6, L (8 , n ) = 6 or 4. Further, for each last digit that has two possible values, the possible values will be present equally as often. Now define S ( d ) such that S (0)=0 and for 1 ≤ d ≤ 9, S ( d ) = L ( d, d ) + L ( d, d + 10) + L ( d, d + 20) + L ( d, d + 30) + · · · + L ( d, d + 90), so that the sum we want to calculate becomes S (0) + S (1) + S (2) + · · · + S (9). But by the above calculations all S ( d ) are divisible by 10, so their sum is divisible by 10, which means its last digit is 0 . 1 1