Difference between revisions of "2013 AMC 10B Problems/Problem 20"
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The number <math> 2013 </math> is expressed in the form <cmath> 2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!}, </cmath> 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>? | The number <math> 2013 </math> is expressed in the form <cmath> 2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!}, </cmath> 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 19:21, 21 February 2013
Problem
The number 
is expressed in the form ![\[2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!},\]](http://latex.artofproblemsolving.com/1/8/5/18571a68f56112253f4bcf42a6895f65bbd41904.png)
where 
and 
are positive integers and 
is as small as possible. What is 
?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}}\ 4\qquad\textbf{(E)}\ 5$ (Error compiling LaTeX. Unknown error_msg)
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 ![\[\frac{61!*19!*11!}{59!*20!*10!},\]](http://latex.artofproblemsolving.com/f/e/a/fea0315109e129513f17641986d9a07aa4bbde38.png)
