HMMT 十一月 2012 · 冲刺赛 · 第 7 题
HMMT November 2012 — Guts Round — Problem 7
题目详情
英文原题
- [ 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
英文解析
- [ 7 ]
Answer: 335
2 na = 1 + 3 + 3 + ... + 3
2 3 n +1 n
3 a = 3 + 3 + 3 + ... + 3
n +1 n
3 a = 3 − 1 + Sn nn +1
3 − 1
a =
2 nn +1
3 − 1
n +1
Because gcd(2 , 7) = 1, to demonstrate that 7 | a = is equivalent to showing that 3 − 1 ≡ 0
2 nmod 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
possible terms.6