HMMT 二月 2001 · 代数 · 第 5 题

HMMT February 2001 — Algebra — Problem 5

Find the 6-digit number beginning and ending in the digit 2 that is the product of three consecutive even integers. 1 2 3 100

解析

Find the 6-digit number beginning and ending in the digit 2 that is the product of three consecutive even integers. Solution: Because the last digit of the product is 2, none of the three consecutive even integers end in 0. Thus they must end in 2, 4, 6 or 4, 6, 8, so they must end in 4, 6, 8 since 2 · 4 · 6 3 does not end in 2. Call the middle integer n . Then the product is ( n − 2) n ( n + 2) = n − 4 n , √ √ √ √ 3 3 3 3 3 3 so n > 200000 = 200 · 10 ≈ 60, but clearly n < 300000 = 300 · 10 < 70. Thus 3 n = 66, and the product is 66 − 4 · 66 = 287232 . 1 2 3 100