PUMaC 2021 · 数论（A 组） · 第 1 题
PUMaC 2021 — Number Theory (Division A) — Problem 1
题目详情
- Compute the remainder when 2 + 3 + 5 is divided by 30.
解析
- = v ((2 ) + 1 ) , which in turn equals v ( a ) + v ( ) = p p p i ( p )+ k a i ( p )+ k − 1 2 v ( a ) − v ( a ) = ( k + 2) a , which gives us our desired. Finally, notice that that p p i ( p )+ k i ( p )+ k − 1 i ( p ) 513 a = 2 + 1 , by trying the first two values. Notice that for each prime p that divides a , if 3 3 j is the index so p first divides a , it follows that the first index k where the power is up by 3 j is so that ( k + 1 − j ) = 3(4 − j ) , or that k = 12 − 3 j − 1 + j = 11 − 2 j. Noticing that a = 3 , 0 ′ divisible by 3 , we therefore have our index being 11 = m and therefore m , by a similar logic, equals 35 . 4