Mock AIME 2 Pre 2005 Problems/Problem 1

Problem

Compute the largest integer $k$ such that $2004^k$ divides $2004!$ .

Solution

Note that $2004 = 2^2 \cdot 3 \cdot 167$ . We focus on the large prime $167$ as the powers of $2$ and $3$ in the prime factorization of $2004!$ are going to be much higher. The largest power of $167$ that divides $2004!$ is $\tfrac{2004}{167} = \boxed{012}$ , the answer.

-MP8148

Mock AIME 2 Pre 2005 (Problems, Source)
Preceded by First question	Followed by Problem 2
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Mock AIME 2 Pre 2005 Problems/Problem 1

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Solution

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