# Difference between revisions of "1998 AIME Problems/Problem 1"

## Problem

For how many values of $k$ is $12^{12}$ the least common multiple of the positive integers $6^6$, $8^8$, and $k$?

## Solution

It is evident that $k$ has only 2s and 3s in its prime factorization, or $k = 2^a3^b$.

• $6^6 = 2^6\cdot3^6$
• $8^8 = 2^{24}$
• $12^{12} = 2^{24}\cdot3^{12}$

The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. $[6^6,8^8] = 2^{24}3^6$. Therefore $12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}$, and $b = 12$. Since $0 \le a \le 24$, there are $\boxed{025}$ values of $k$.