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Difference between revisions of "2013 AMC 10B Problems/Problem 20"

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==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>
 +
== See also ==
 +
{{AMC10 box|year=2013|ab=B|num-b=19|num-a=21}}

Revision as of 17:05, 27 March 2013

Problem

The number $2013$ is expressed in the form \[2013=\frac{a_1!a_2!\cdots a_m!}{b_1!b_2!\cdots b_n!},\] where $a_1\ge a_2\ge\cdots\ge a_m$ and $b_1\ge b_2\ge\cdots\ge b_n$ are positive integers and $a_1+b_1$ is as small as possible. What is $|a_1-b_1|$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}}\ 4\qquad\textbf{(E)}\ 5$ (Error compiling LaTeX. Unknown error_msg)

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 \[\frac{61!*19!*11!}{59!*20!*10!},\]

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
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