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HMMT 二月 2002 · 代数 · 第 5 题

HMMT February 2002 — Algebra — Problem 5

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Find the greatest common divisor of the numbers 2002 + 2 , 2002 + 2 , 2002 + 2 , . . . .
解析
  1. Find the greatest common divisor of the numbers 2002 + 2 , 2002 + 2 , 2002 + 2 , . . . . 2 2 Solution: 6 . Notice that 2002+2 divides 2002 − 2 , so any common divisor of 2002+2 2 2 2 2 and 2002 + 2 must divide (2002 + 2) − (2002 − 2 ) = 6. On the other hand, every number n in the sequence is even, and the n th number is always congruent to 1 + 2 ≡ 0 modulo 3. Thus, 6 divides every number in the sequence.