2013 AMC 10B Problem 20

Factor: $2013 = 3 \cdot 11 \cdot 61$.

Lower bound. The prime $61$ must divide the numerator, and $61 \mid k!$ only when $k \ge 61$, so $a_1 \ge 61$. But then $a_1!$ is divisible by the prime $59$ (and by $53, 47, \dots$), which does not divide $2013$, so $59$ must be cancelled by the denominator: some $b_j!$ is divisible by $59$, forcing $b_1 \ge 59$. Hence $a_1 + b_1 \ge 61 + 59 = 120$.

Construction. Try $a_1 = 61$, $b_1 = 59$: $\frac{61!}{59!} = 61 \cdot 60 = 3660$, and $\frac{3660}{2013} = \frac{60}{33} = \frac{20}{11}$. We still need to divide by $20$ and multiply by $11$, which factorials do easily:
$$2013 = \frac{61! \cdot 19! \cdot 11!}{59! \cdot 20! \cdot 10!},$$
since $\frac{19!}{20!} = \frac{1}{20}$ and $\frac{11!}{10!} = 11$. The ordering conditions hold, and $a_1 + b_1 = 120$ is minimal.

Therefore $|a_1 - b_1| = 61 - 59 = 2$.

The answer is $\boxed{\textbf{(B)}\ 2}$.

Ratios of factorials are products of consecutive integers, so the largest prime in the target dictates the largest factorial, and every unwanted prime just below it must be cancelled by a denominator factorial of nearly the same size. Prove the bound with the largest prime and the next prime down, then build an explicit expression to show it is attained.

• Taking b_1 = 60 (so 61!/60! = 61) without checking whether a smaller b_1 still lets the leftover primes cancel.
• Assuming $a_1 = 2013$ or otherwise ignoring the requirement that $a_1 + b_1$ be minimized.