2013 AMC 12B Problems/Problem 15
Problem
The number is expressed in the form
![$2013 = \frac {a_1!a_2!...a_m!}{b_1!b_2!...b_n!}$](http://latex.artofproblemsolving.com/6/6/9/6690a904c827dd9d3643929e24f32e7eea1074e5.png)
where and
are positive integers and
is as small as possible. What is
?
Solution
The prime factorization of is
. To have a factor of
in the numerator,
must equal
. Now we notice that there can be no prime
which is not a factor of 2013 such that
because this prime will not be represented in the denominator, but will be represented in the numerator. The highest
less than
is
, so there must be a factor of
in the denominator. It follows that
, so the answer is
, which is
. One possible way to express
is
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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All AMC 12 Problems and Solutions |