HMMT 十一月 2014 · 冲刺赛 · 第 19 题
HMMT November 2014 — Guts Round — Problem 19
题目详情
英文原题
- [ 11 ] Let a sequence { a } be defined by a = 2 , a = 2 , and a = a a for n ≥ 1. Then 0 1 n +1 nn =0 n − 1
sequence of remainders when a , a , a , · · · are divided by 2014 is eventually periodic with some minimal
0 1 2
period p (meaning that a = a for all sufficiently large integers m , and p is the smallest suchm m + ppositive integer). Find p .
解析
英文解析
- [ 11 ] Let a sequence { a } be defined by a = 2 , a = 2 , and a = a a for n ≥ 1. Then 0 1 n +1 nn =0 n − 1
sequence of remainders when a , a , a , · · · are divided by 2014 is eventually periodic with some minimal
0 1 2
period p (meaning that a = a for all sufficiently large integers m , and p is the smallest suchm m + ppositive integer). Find p .
Answer: 12 Let a = 2 , so notice b = 1 , b = 2 , and b = b + 2 b for n ≥ 1, so bynbn 1 2 n +1 n n − 1
n − 1
n − 1 2
inspection b = 2 for all n ; thus a = 2 . 2014 = 2 · 19 · 53 so we just want to find the lcm ofn nthe eventual periods of 2 mod ord (2) and ord (2). These orders divide 18 and 52 respectively, andn
19 53
we can manually check ord (2) = 6 and ord (2) = 12. The lcm of these is 12, so to show the answer
9 13
is 12, it suffices to show that 13 | ord (2). This is true since 2 6 ≡ 1 (mod 53). So, the answer is 12 .4
53