2013 AMC 12B Problems/Problem 15
- The following problem is from both the 2013 AMC 12B #15 and 2013 AMC 10B #20, so both problems redirect to this page.
Problem
The number 
is expressed in the form
,
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 and to minimize 
must equal . Now we notice that there can be no prime 
which is not a factor of such that 
because this prime will not be canceled out in the denominator, and will lead to an extra factor in the numerator. The highest prime less than is 
, so there must be a factor of in the denominator. It follows that 
(to minimize 
as well), so the answer is 
. One possible way to express with 
is ![\[2013 = \frac{61!\cdot19!\cdot11!}{59!\cdot20!\cdot10!}.\]](http://latex.artofproblemsolving.com/d/e/f/def17d519ef8334692dce639518195ebe0c6da8b.png)
Video Solution
https://youtube.com/FvscTObzpwA
~IceMatrix
Video Solution 2
~savannahsolver
See also
| 2013 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2013 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
