HMMT 二月 2004 · 冲刺赛 · 第 2 题

英文解析

2. Now, in general, let p = min( { p , . . . , p } \ { 2 } ). Suppose 1 < j < p . By induction,
   1 nwe have a = ( j + 1) a / 2 when j is odd, and a = a / 2 when j is even. So a 6 = a forj 1 j 1 i 1
   all 1 < j < p . It follows that a = a / 2 p . Then, again using induction, we get for allp 1
   nonnegative integers k that a = a if k is even, and a = ( p + k + 1) a if k is odd.
   p + k p p + k p
   Clearly, a 6 = a and p + k + 1 6 = 2 p when k is odd (the left side is odd, and the rightp 1
   side even). It follows that a = a for no j > 1. Finally, when a = 2, we can checkj 1 1
   inductively that a = j + 1 for j odd and a = 1 for j even.
   j j
   So our answer is just the number of odd elements in S . There are 9 odd prime numberssmaller than 30, so the answer is 2 = 512.9