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

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<math>x \cdot \frac{11}{12} \cdot \frac{5}{6} \cdot \frac{1}{2}  \cdot \frac{7}{12} \cdot \frac{1}{2} \cdot \frac{5}{12} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{6} \cdot \frac{1}{12}</math>.  Now, the minimal <math>x</math> 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 <math>12^{\text{th}}</math> pirate receives, as he receives <math>\frac{12}{12} = 1 =</math> all of what is remaining.   
 
<math>x \cdot \frac{11}{12} \cdot \frac{5}{6} \cdot \frac{1}{2}  \cdot \frac{7}{12} \cdot \frac{1}{2} \cdot \frac{5}{12} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{6} \cdot \frac{1}{12}</math>.  Now, the minimal <math>x</math> 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 <math>12^{\text{th}}</math> pirate receives, as he receives <math>\frac{12}{12} = 1 =</math> 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, <math>11 \cdot 5 \cdot 7 \cdot 5 = \boxed{\textbf{(D) }1925}</math>.
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Thus, we know the denominator is canceled out, so the number of gold coins received is going to be the product of the numerators, <math>11 \cdot 5 \cdot 7 \cdot 5 = \boxed{\textbf{(D) }1925}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 15:22, 23 April 2017

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

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


$x \cdot \frac{11}{12} \cdot \frac{5}{6} \cdot \frac{1}{2}  \cdot \frac{7}{12} \cdot \frac{1}{2} \cdot \frac{5}{12} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{6} \cdot \frac{1}{12}$. 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 canceled out, so the number of gold coins received is going to be the product of the numerators, $11 \cdot 5 \cdot 7 \cdot 5 = \boxed{\textbf{(D) }1925}$.

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

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