Difference between revisions of "2013 AMC 10A Problems/Problem 21"

Line 22: Line 22:
 
{{AMC10 box|year=2013|ab=A|num-b=20|num-a=22}}
 
{{AMC10 box|year=2013|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}}
 
{{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Revision as of 12:08, 4 July 2013

Problem

A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?


$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850$

Solution

Let $x$ be the number of coins. After the $k^{\text{th}}$ pirate takes his share, $\frac{12-k}{12}$ of the original amount is left. Thus, we know that

$x * \frac{11}{12} * \frac{10}{12} * \frac{9}{12} * \frac{8}{12} * \frac{7}{12} * \frac{6}{12} * \frac{5}{12} * \frac{4}{12} * \frac{3}{12} * \frac{2}{12} * \frac{1}{12}$ must be an integer. Simplifying, we get


$x * \frac{11}{12} * \frac{5}{6} * \frac{1}{2}  * \frac{7}{12} * \frac{1}{2} * \frac{5}{12} * \frac{1}{3} * \frac{1}{4} * \frac{1}{6} * \frac{1}{12}$. Now, the minimal x is the denominator of this fraction multiplied out, obviously. We mentioned before that this product must be an integer. Specifically, it is an integer and it is the amount that the $12^{\text{th}}$ pirate receives, as he receives $\frac{12}{12} = 1 =$ all of what is remaining.

Thus, we know the denominator is cancelled out, so the number of gold coins received is going to be the product of the numerators, $11 * 5 * 7 * 5 = \boxed{\textbf{(D) }1925}$.


See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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
2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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