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

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

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

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

@@ Line 11: / Line 11: @@
 == See also ==
 {{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]]
 {{MAA Notice}}

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

Revision as of 22:42, 22 November 2013

Problem

Solution

See also