1998 AIME Problems/Problem 1

Problem

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

Solution

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

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

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

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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Solution

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