HMMT 十一月 2016 · THM 赛 · 第 4 题

HMMT November 2016 — THM Round — Problem 4

A positive integer is written on each corner of a square such that numbers on opposite vertices are relatively prime while numbers on adjacent vertices are not relatively prime. What is the smallest possible value of the sum of these 4 numbers?

解析

A positive integer is written on each corner of a square such that numbers on opposite vertices are relatively prime while numbers on adjacent vertices are not relatively prime. What is the smallest possible value of the sum of these 4 numbers? Proposed by: Eshaan Nichani Answer: 60 Two opposite vertices are relatively prime, but they both share a factor with their common neighbor. So that common neighbor must have two prime factors. So each of the 4 numbers has two prime factors, which are not shared with the opposite vertex. Moreover, it suffices to choose the vertices to be the numbers ab, bc, cd, da for some prime numbers a, b, c, d . It’s clear that we should choose them to be the smallest primes 2 , 3 , 5 , 7 in some order. The order that minimizes the sum of all of the numbers gives 14 , 10 , 15 , 21 for a sum of 60.