Difference between revisions of "2013 AMC 12B Problems/Problem 15"
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− | The prime factorization of <math> 2013 </math> is <math> 61*11*3 </math>. To have a factor of <math>61</math> in the numerator, <math>a_1</math> must equal <math>61</math>. Now we notice that there can be no prime <math>p</math> which is not a factor of 2013 such that <math> b_1<p<61</math> because this prime will not be represented in the denominator, but will be represented in the numerator. The highest <math> p </math> less than <math>61</math> is <math>59</math>, so there must be a factor of <math>59</math> in the denominator. It follows that <math>b_1 = 59</math>, so the answer is <math>|61-59| | + | The prime factorization of <math> 2013 </math> is <math> 61*11*3 </math>. To have a factor of <math>61</math> in the numerator and to minimize <math>a_1,</math> <math>a_1</math> must equal <math>61</math>. Now we notice that there can be no prime <math>p</math> which is not a factor of 2013 such that <math> b_1<p<61,</math> because this prime will not be represented in the denominator, but will be represented in the numerator. The highest <math> p </math> less than <math>61</math> is <math>59</math>, so there must be a factor of <math>59</math> in the denominator. It follows that <math>b_1 = 59</math> (to minimize <math>b_1</math> as well), so the answer is <math>|61-59| = \boxed{\textbf{(B) }2}</math>. One possible way to express <math> 2013 </math> is <cmath> 2013 = \frac{61!*19!*11!}{59!*20!*10!}, </cmath> |
== See also == | == See also == |
Revision as of 02:37, 20 January 2014
- 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
![$2013 = \frac {a_1!a_2!...a_m!}{b_1!b_2!...b_n!}$](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 and to minimize
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
(to minimize
as well), so the answer is
. One possible way to express
is
See also
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.