AIME 1998 · 第 1 题

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

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