Difference between revisions of "2013 AMC 12B Problems/Problem 15"
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==Problem== | ==Problem== | ||
− | The number <math>2013</math> is expressed in the form <br \> <center> <math>2013 = \frac {a_1!a_2!...a_m!}{b_1!b_2!...b_n!}</math>,</center><br />where <math>a_1 \ge a_2 \ge | + | The number <math>2013</math> is expressed in the form <br \> <center> <math>2013 = \frac {a_1!a_2!...a_m!}{b_1!b_2!...b_n!}</math>,</center><br />where <math>a_1 \ge a_2 \ge \cdots \ge a_m</math> and <math>b_1 \ge b_2 \ge \cdots \ge b_n</math> are positive integers and <math>a_1 + b_1</math> is as small as possible. What is <math>|a_1 - b_1|</math>? |
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math> | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math> | ||
==Solution== | ==Solution== | ||
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|</math>, which is <math>\boxed{\textbf{(B) }2}</math>. One possible way to express <math> 2013 </math> is <cmath> \frac{61!*19!*11!}{59!*20!*10!}, </cmath> | 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|</math>, which is <math>\boxed{\textbf{(B) }2}</math>. One possible way to express <math> 2013 </math> is <cmath> \frac{61!*19!*11!}{59!*20!*10!}, </cmath> |
Revision as of 16:44, 22 February 2013
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