2013 AMC 10B Problems/Problem 20

Problem

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 $|a_1-b_1|$ ?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D}}\ 4\qquad\textbf{(E)}\ 5$ (Error compiling LaTeX. Unknown error_msg)

Solution

The prime factorization of $2013$ is $61*11*3$ . To have a factor of $61$ in the numerator, $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is not a factor of 2013 such that $b_1<p<61$ because this prime will not be represented in the denominator, but will be represented in the numerator. The highest $p$ less than $61$ is $59$ , so there must be a factor of $59$ in the denominator. It follows that $b_1 = 59$ , so the answer is $|61-59|$ , which is $\boxed{\textbf{(B) }2}$ . One possible way to express $2013$ is $\frac{61!*19!*11!}{59!*20!*10!},$

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2013 AMC 10B Problems/Problem 20

Problem

Solution