# Problem #1816

 1816 The number $2013$ is expressed in the form $$2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!},$$ where $a_1\ge a_2\ge\cdots\ge a_m$ and $b_1\ge b_2\ge\cdots\ge b_n$ are positive integers and $a_1+b_1$ is as small as possible. What is $|a_1-b_1|$? $\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ This problem is copyrighted by the American Mathematics Competitions.
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