HMMT 十一月 2009 · GEN2 赛 · 第 9 题

HMMT November 2009 — GEN2 Round — Problem 9

[ 6 ] Five guys each have a positive integer (the integers are not necessarily distinct). The greatest common divisor of any two guys’ numbers is always more than 1, but the greatest common divisor of all the numbers is 1. What is the minimum possible value of the product of the numbers?

解析

[ 6 ] Five guys each have a positive integer (the integers are not necessarily distinct). The greatest common divisor of any two guys’ numbers is always more than 1, but the greatest common divisor of all the numbers is 1. What is the minimum possible value of the product of the numbers? Answer: 32400 Let ω ( n ) be the number of distinct prime divisors of a number. Each of the guys’ numbers must have ω ( n ) ≥ 2, since no prime divides all the numbers. Therefore, if the answer has e e e 1 2 k 2 prime factorization p p . . . p , then e + e + . . . + e ≥ 10. If p divided any of the guys’ numbers, 1 2 k 1 2 k we could divide their number by p to reduce the product. Therefore we may assume e ≤ 4 for each i 4 4 2 i , so the smallest possible product is 2 3 5 . This bound is achievable: give the guys the numbers 10 , 6 , 6 , 6 , and 15.