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Difference between revisions of "2013 AMC 12A Problems/Problem 17"

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== Problem 17 ==
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A group of <math> 12 </math> pirates agree to divide a treasure chest of gold coins among themselves as follows. The <math> k^\text{th} </math> pirate to take a share takes <math> \frac{k}{12} </math> 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 <math> 12^{\text{th}} </math> pirate receive?
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<math> \textbf{(A)} \ 720 \qquad  \textbf{(B)} \ 1296 \qquad  \textbf{(C)} \ 1728 \qquad  \textbf{(D)} \ 1925 \qquad  \textbf{(E)} \ 3850 </math>
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==Solution==
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The first pirate takes <math>\frac{1}{12}</math> of the <math>x</math> coins, leaving <math>\frac{11}{12} x</math>.
 
The first pirate takes <math>\frac{1}{12}</math> of the <math>x</math> coins, leaving <math>\frac{11}{12} x</math>.
  
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We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, <math>x</math> is the denominator, leaving <math>1925</math> coins for the twelfth pirate.
 
We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, <math>x</math> is the denominator, leaving <math>1925</math> coins for the twelfth pirate.
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}}

Revision as of 18:44, 22 February 2013

Problem 17

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

The first pirate takes $\frac{1}{12}$ of the $x$ coins, leaving $\frac{11}{12} x$.

The second pirate takes $\frac{2}{12}$ of the remaining coins, leaving $\frac{10}{12}*\frac{11}{12}*x$.

Continuing this pattern, the eleventh pirate must take $\frac{11}{12}$ of the remaining coins after the first ten pirates have taken their share, which leaves $\frac{11!}{12^{12}}*x$. The twelfth pirate takes all of this.


Note that

$12^{12} = (2^2 * 3)^{12} = 2^{24} * 3^{12}$

$11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2$


All the 2s and 3s cancel out of $11!$, leaving

$11 * 5 * 7 * 5 = 1925$

in the numerator.


We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, $x$ is the denominator, leaving $1925$ coins for the twelfth pirate.


See also

2013 AMC 12A (Problems • Answer Key • Resources)
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
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