2013 AMC 12B Problem 15

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

Factor

2013 = 3\cdot 11\cdot 61;

its largest prime is

61

Some

a_i!

must reach

61,

so

a_1 \ge 61;

but

61!

also contains the prime

59,

which must be cancelled from the denominator

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.

Problem 15 in other years