HMMT 十一月 2023 · GEN 赛 · 第 8 题

HMMT November 2023 — GEN Round — Problem 8

Mark writes the expression d for each positive divisor d of 8! on the board. Seeing that these √ expressions might not be worth points on HMMT, Rishabh simplifies each expression to the form a b , where a and b are integers such that b is not divisible by the square of a prime number. (For example, √ √ √ √ √ √ 20 , 16 , and 6 simplify to 2 5 , 4 1 , and 1 6 , respectively.) Compute the sum of a + b across all expressions that Rishabh writes.

解析

Mark writes the expression d for each positive divisor d of 8! on the board. Seeing that these √ expressions might not be worth points on HMMT, Rishabh simplifies each expression to the form a b , where a and b are integers such that b is not divisible by the square of a prime number. (For example, √ √ √ √ √ √ 20 , 16 , and 6 simplify to 2 5 , 4 1 , and 1 6 , respectively.) Compute the sum of a + b across all expressions that Rishabh writes. Proposed by: Pitchayut Saengrungkongka Answer: 3480 √ √ ∑ Solution: Let n simplify to a b . Notice that both a and b are multiplicative. Thus, a n n n n d d | n ∑ and b are multiplicative. d d | n ∑ ∑ l ⌊ l/ 2 ⌋ 2 { l/ 2 } We consider the sum a and b . Notice that for d = p , a = p and b = p , so k k d d d d d | p d | p ( ) ( k +1) / 2 ∑ ∑ p − 1 ( p + 1)( k + 1) a = 2 and b = d d p − 1 2 k k d | p d | p for odd k , while ( ) ( k +2) / 2 k/ 2 ∑ ∑ p + p − 2 ( p + 1) k a = and b = + 1 d d p − 1 2 k k d | p d | p for even k . 7 2 Notice 8! = 2 · 3 · 5 · 7 , so ( ) ( ) ∑ 2(16 − 1) 9 + 3 − 2 a = (1 + 1) (1 + 1) = 30 · 5 · 2 · 2 = 600 d 2 − 1 3 − 1 d | 8! and ( ) ( ) ∑ 3 · 8 4 · 2 b = 1 + (1 + 5) (1 + 7) = 12 · 5 · 6 · 8 = 2880 , d 2 2 d | 8! so the sum of a + b would be 600 + 2880 = 3480 . d d