# 2013 AMC 12B Problems/Problem 15

The following problem is from both the 2013 AMC 12B #15 and 2013 AMC 10B #20, so both problems redirect to this page.

## Problem

The number $2013$ is expressed in the form $2013 = \frac {a_1!a_2!...a_m!}{b_1!b_2!...b_n!}$,

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 $|a_1 - b_1|$? $\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

## Solution

The prime factorization of $2013$ is $61\cdot11\cdot3$. To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$. Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1 because this prime will not be canceled out in the denominator, and will lead to an extra factor in the numerator. The highest prime less than $61$ is $59$, so there must be a factor of $59$ in the denominator. It follows that $b_1 = 59$ (to minimize $b_1$ as well), so the answer is $|61-59| = \boxed{\textbf{(B) }2}$. One possible way to express $2013$ with $(a_1, b_1) = (61, 59)$ is $$2013 = \frac{61!\cdot19!\cdot11!}{59!\cdot20!\cdot10!}.$$

## Video Solution

~IceMatrix

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