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 1

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 $\boxed{25}$ values of $k$ .

Solution 2

We want the number of $k$ such that $\operatorname{lcm}(6^6,8^8,k)=12^12$ . Using $\operatorname{lcm}$ properties, this is $\operatorname{lcm}(\operatorname{lcm}(2^6\cdot 3^6,2^24),k)=2^24 \cdot 3^12$ , or $\operatorname{lcm}(2^24\cdot 3^6,k)=2^24\cdot 3^12$ . At this point, we realize that $k=2^a\cdot3^b$ , as any other prime factors would be included in the $\operatorname{lcm}$ . Also, $b=12$ (or the power of $3$ in the $\operatorname{lcm}$ wouldn't be $12$ ) and $0\le a\le 24$ (or the power of $2$ in the $\operatorname{lcm}$ would be $a$ and not $24$ ). Therefore, $a$ can be any integer from $0$ to $24$ , for a total of $25$ values of $a$ and $\boxed{025}$ values of $k$ .

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=2899

~ pi_is_3.14

1998 AIME (Problems • Answer Key • Resources)
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1998 AIME Problems/Problem 1

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Problem

Solution 1

Solution 2

Video Solution by OmegaLearn

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