HMMT 二月 2004 · 冲刺赛 · 第 38 题
HMMT February 2004 — Guts Round — Problem 38
题目详情
英文原题
- [15] Let S = { p p · · · p | p , p , . . . , p are distinct primes and p , . . . , p < 30 } . As-
1 2 n 1 2 n 1 nsume 1 is in S . Let a be an element of S . We define, for all positive integers n :
a = a / ( n + 1) if a is divisible by n + 1;1
n +1 n na = ( n + 2) a if a is not divisible by n + 1 .
n +1 n n
How many distinct possible values of a are there such that a = a for infinitely many
1 j 1
j ’s?
解析
英文解析
- Let S = { p p · · · p | p , p , . . . , p are distinct primes and p , . . . , p < 30 } . Assume
1 2 n 1 2 n 1 n
1 is in S . Let a be an element of S . We define, for all positive integers n :
a = a / ( n + 1) if a is divisible by n + 1;1
n +1 n na = ( n + 2) a if a is not divisible by n + 1 .
n +1 n n
How many distinct possible values of a are there such that a = a for infinitely many
1 j 1
j ’s?
Solution: 512
If a is odd, then we can see by induction that a = ( j +1) a when j is even and a = a
1 j 1 j 1
when j is odd (using the fact that no even j can divide a ). So we have infinitely manyj ’s for which a = a .1
j 1
If a > 2 is even, then a is odd, since a = a / 2, and a may have only one factor of
1 2 2 1 1