2013 AMC 12B Problem 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|\,?

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Solution

Since

2013 = 3\cdot 11\cdot 61,

the numerator needs a factorial at least

61!

to supply the prime

61,

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.

Tagged: prime factorization · factorial · bounding to limit cases