# 2000 PMWC Problems/Problem I1

## Problem

$y$ is a number that has $8$ different factors (including the number $1$ and itself). What is the smallest possible value of $y$?

## Solution

The formula for the number of factors is found by adding one to each exponent in the prime factorization, then finding the product of these exponents. In this case, we want the smallest possible number with 8 total factors.

There are three distinct "configurations" of prime factorizations that will yield 8 factors. They are $p^7$, $p^3 * q^1$, and $p^1 * q^1 * r^1$, since $(7+1) = 8$, $(3+1)(1+1)=8$, and $(1+1)(1+1)(1+1) = 8$, respectively. (In the previously listed prime configurations, the letters $p$, $q$, and $r$, stand for distinct primes.)

For each of these cases, the smallest possible number that satisfies this configuration is computed. 2 is the smallest prime, so 2 is always used as the base for the prime with the largest exponent. The next largest exponent should have a base of 3, then 5, then 7, and so on. In the first case, we have $2^7 = 128$, in the second case, we have $2^3 * 3^1 = 24$, and in the third case, we have $2^1 * 3^1 * 5^1 = 30$.

Since the second case yields the lowest number, $24$ is the answer. You can check by listing out the factors: 1, 2, 3, 4, 6, 8, 12, and 24.