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

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==Solution==
 
==Solution==
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The prime factorization of <math> 2013 </math> is <math> 61\cdot11\cdot3 </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 <math>2013</math> 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> with <math>(a_1, b_1) = (61, 59)</math> is  <cmath> 2013 = \frac{61!\cdot19!\cdot11!}{59!\cdot20!\cdot10!}. </cmath>
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The prime factorization of <math> 2013 </math> is <math> 61\cdot11\cdot3 </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 <math>2013</math> such that <math> b_1<p<61,</math> because this prime will not be canceled out in the denominator, and will lead to an extra factor 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> with <math>(a_1, b_1) = (61, 59)</math> is  <cmath> 2013 = \frac{61!\cdot19!\cdot11!}{59!\cdot20!\cdot10!}. </cmath>
  
 
== See also ==
 
== See also ==

Revision as of 14:52, 20 November 2016

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 $2013$ is expressed in the form

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


where $a_1 \ge a_2 \ge \cdots \ge a_m$ 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 $|a_1 - b_1|$?

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

Solution

The prime factorization of $2013$ is $61\cdot11\cdot3$. To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ 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 canceled out in the denominator, and will lead to an extra factor 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 $b_1 = 59$ (to minimize $b_1$ as well), so the answer is $|61-59| = \boxed{\textbf{(B) }2}$. One possible way to express $2013$ with $(a_1, b_1) = (61, 59)$ is \[2013 = \frac{61!\cdot19!\cdot11!}{59!\cdot20!\cdot10!}.\]

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
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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