HMMT 十一月 2012 · 冲刺赛 · 第 7 题
HMMT November 2012 — Guts Round — Problem 7
题目详情
- [ 7 ] Consider the sequence given by a = 1 , a = 1 + 3 , a = 1 + 3 + 3 , a = 1 + 3 + 3 + 3 , . . . . Find 0 1 2 3 the number of terms among a , a , a , . . . , a that are divisible by 7. 0 1 2 2012
解析
- [ 7 ] Answer: 335 2 n a = 1 + 3 + 3 + ... + 3 n 2 3 n +1 3 a = 3 + 3 + 3 + ... + 3 n n +1 3 a = 3 − 1 + S n n n +1 3 − 1 a = n 2 n +1 3 − 1 n +1 Because gcd(2 , 7) = 1, to demonstrate that 7 | a = is equivalent to showing that 3 − 1 ≡ 0 n 2 mod 7 n +1 However 3 − 1 = 0 mod 7 = ⇒ n + 1 ≡ 0 mod 6, hence the values of n ≤ 2012 are 5, 11, 17, ..., 2009, which is in total 2009 + 1 = 335 6 possible terms.