The number 2013 is expressed in the form 2013 = (a_1! a_2!\cdots a_m!)/(b_1! b_2!\cdots b_n!), where a_1 ≥ a_2 ≥ \cdots ≥ a_m and b_1 ≥ b_2 ≥ \cdots ≥ b_n are positive integers and a_1 + b_1 is as small as possible. What is |a_1 - b_1| ?

2013 AMC 12B Problem 15 — Solution

2013 AMC 12B Problem 15

Below is the professionally curated solution for Problem 15 of the 2013 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12B solutions, or check the answer key.

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Concepts:prime factorization factorial bounding to limit cases

Difficulty rating: 1840

15.

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 $|a_1 - b_1|\,?$

$1$

$2$

$3$

$4$

$5$

Solution:

Since $2013 = 3\cdot 11\cdot 61,$ the numerator needs a factorial at least $61!$ to supply the prime $61,$ so $a_1 \ge 61.$ But $61!$ also has a factor of $59,$ which $2013$ does not, so the denominator needs $b_1 \ge 59.$ Thus $a_1 + b_1 \ge 120,$ attained by $a_1 = 61,$ $b_1 = 59$ via $2013 = \dfrac{61!\,11!\,3!}{59!\,10!\,5!}.$ Then $|a_1 - b_1| = 2.$ Thus, the correct answer is B.

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