HMMT 二月 2021 · 冲刺赛 · 第 8 题

HMMT February 2021 — Guts Round — Problem 8

[9] Compute the product of all positive integers b ≥ 2 for which the base b number 111111 has exactly b b distinct prime divisors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT Spring 2021, March 06, 2021 — GUTS ROUND Organization Team Team ID#

解析

[9] Compute the product of all positive integers b ≥ 2 for which the base b number 111111 has exactly b b distinct prime divisors. Proposed by: Esha Bhatia Answer: 24 Solution: Notice that this value, in base b , is 6 b − 1 2 2 = ( b + 1)( b − b + 1)( b + b + 1) . b − 1 2 2 This means that, if b satisfies the problem condition, ( b + 1)( b − b + 1)( b + b + 1) > p . . . p , where 1 b p is the i th smallest prime. i 2 2 We claim that, if b ≥ 7, then p . . . p > ( b + 1)( b − b + 1)( b + b + 1) . This is true for b = 7 by 1 b calculation, and can be proven for larger b by induction and the estimate p ≥ i . i All we have to do is to check b ∈ 2 , 3 , 4 , 5 , 6 . Notice that for b = 6 , the primes cannot include 2 , 3 6 6 − 1 and hence we want to be divisible product of 6 primes the smallest of which is 5. However, 5 6 6 − 1 5 · 7 · · · · 17 > , and by checking we rule out 5 too. All that is left is { 2 , 3 , 4 } , all of which work, 5 giving us an answer of 24.