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Difference between revisions of "2013 AMC 12B Problems/Problem 15"

(Created page with "==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...")
 
Line 3: Line 3:
 
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 ... \ge a_m</math> and <math>b_1 \ge b_2 \ge ... \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 <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 ... \ge a_m</math> and <math>b_1 \ge b_2 \ge ... \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==
 +
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 15:43, 22 February 2013

Problem

the number $2013$ is expressed in the form

$2013 = \frac {a_1!a_2!...a_m!}{b_1!b_2!...b_n!}$,


where $a_1 \ge a_2 \ge ... \ge a_m$ and $b_1 \ge b_2 \ge ... \ge b_n$ are positive integers and $a_1 + b_1$ is as small as possible. What is $|a_1 - b_1|$?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution

The prime factorization of $2013$ is $61*11*3$. To have a factor of $61$ in the numerator, $a_1$ must equal $61$. Now we notice that there can be no prime $p$ which is not a factor of 2013 such that $b_1<p<61$ because this prime will not be represented in the denominator, but will be represented in the numerator. The highest $p$ less than $61$ is $59$, so there must be a factor of $59$ in the denominator. It follows that $b_1 = 59$, so the answer is $|61-59|$, which is $\boxed{\textbf{(B) }2}$. One possible way to express $2013$ is \[\frac{61!*19!*11!}{59!*20!*10!},\]

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