Question

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 $|a_1 - b_1|?$

$1$

$2$

$3$

$4$

$5$

Answer: B

Concepts:factorial prime factorization

Difficulty rating: 2060

Solution:

The prime factorization is $2013=3\cdot11\cdot61$ , so the numerator must contain a factor of $61$ . Hence $a_1\ge61$ .

But $61!$ also contains the prime factor $59$ , which is not in $2013$ , so the denominator must contain a factor of $59$ . Hence $b_1\ge59$ .

The lower bound $a_1+b_1\ge120$ is attainable because $2013=\frac{61!\,11!\,3!}{59!\,10!\,5!}$ .

Thus $|a_1-b_1|=61-59=2$ , and the correct answer is B .

2013 AMC 10B Problem 20

Answer: B

Difficulty rating: 2060

Solution:

Problem 20 in Other Years