2013 AMC 10B Problem 20

Problem 20 of 25HarderNumber Theory Arithmetic

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|?

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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 .

Tagged: factorial · prime factorization

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