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 $k$ .

1998 AIME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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1998 AIME Problems/Problem 1

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