Question

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

Answer

Answer: 25

Concepts:least common multiple prime factorization

Difficulty rating: 1890

Solution:

Since $12^{12} = 2^{24} 3^{12},$ $6^6 = 2^6 3^6,$ and $8^8 = 2^{24},$ the number $k$ can involve no primes other than $2$ and $3,$ so write $k = 2^a 3^b.$ The least common multiple of the three numbers is then $2^{\max(24,\,a)} \, 3^{\max(6,\,b)}.$

Matching this to $2^{24} 3^{12}$ requires $\max(24, a) = 24,$ i.e. $0 \le a \le 24,$ and $\max(6, b) = 12,$ i.e. $b = 12.$ That gives $25$ choices for $a$ and one for $b,$ so there are $25$ values of $k.$

1998 AIME Problem 1

Answer: 25

Difficulty rating: 1890

Solution:

Problem 1 in Other Years