AIME 1990 · 第 5 题

AIME 1990 — Problem 5

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

Let $n^{}_{}$ be the smallest positive integer that is a multiple of $75_{}^{}$ and has exactly $75_{}^{}$ positive integral divisors, including $1_{}^{}$ and itself. Find $\frac{n}{75}$ .

解析

Solution

The prime factorization of $75 = 3^15^2 = (2+1)(4+1)(4+1)$ . For $n$ to have exactly $75$ integral divisors, we need to have $n = p_1^{e_1-1}p_2^{e_2-1}\cdots$ such that $e_1e_2 \cdots = 75$ . Since $75|n$ , two of the prime factors must be $3$ and $5$ . To minimize $n$ , we can introduce a third prime factor, $2$ . Also to minimize $n$ , we want $5$ , the greatest of all the factors, to be raised to the least power. Therefore, $n = 2^43^45^2$ and $\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}$ .

\textbf{ Note: the smallest integer n which has 75 factors, happens to be a multiple of 75 anyway }

Video Solution by OmegaLearn

https://youtu.be/jgyyGeEKhwk?t=588

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=zlFLzuotaMU