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

(See also)
m (See also)
Line 11: Line 11:
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=B|num-b=14|num-a=16}}
 
{{AMC12 box|year=2013|ab=B|num-b=14|num-a=16}}
{{AMC10 box|year=2013|ab=B|num-b=19|num-a=21}}
+
 
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:42, 22 November 2013

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 $61*11*3$. 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 12B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS