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

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We can compute the amount of factors of <math>2</math>, <math>3</math>, <math>5</math>, etc. but this is not necessary. To minimize this expression, we must take out factors of <math>2</math> and <math>3</math>, since <math>12^{11}=2^{22} \cdot 3^{11}</math>. <math>11!</math> has neither <math>22</math> factors of <math>2</math>, nor <math>11</math> factors of <math>3</math>. That means <math>x</math> will substitute those factors, since the expression must have an integer value. Notice now how the numerator will have no factors of <math>2</math> or <math>3</math>.
+
We can compute the amount of factors of <math>2</math>, <math>3</math>, <math>5</math>, etc. but this is not necessary. To minimize this expression, we must take out factors of <math>2</math> and <math>3</math>, since <math>12^{11}=2^{22} \cdot 3^{11}</math>. <math>11!</math> has neither <math>22</math> factors of <math>2</math>, nor <math>11</math> factors of <math>3</math>. This means that if <math>11!</math> contains <math>a</math> factors of <math>2</math>, then <math>x</math> will contain <math>22-a</math> factors of <math>2</math>.
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 +
 
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Thus, once simplified, the expression will have no factors of <math>2</math>. It will also have no factors of <math>3</math>.
  
  

Revision as of 16:33, 7 August 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 1

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, $11 \cdot 5 \cdot 7 \cdot 5 = \boxed{\textbf{(D) }1925}$.

Solution 2 (Using the answer choices)

Solution $1$ mentioned the expression $x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot ... \cdot \frac{1}{12}$. Note that this is equivalent to $\frac{x \cdot 11!}{12^{11}}$.


We can compute the amount of factors of $2$, $3$, $5$, etc. but this is not necessary. To minimize this expression, we must take out factors of $2$ and $3$, since $12^{11}=2^{22} \cdot 3^{11}$. $11!$ has neither $22$ factors of $2$, nor $11$ factors of $3$. This means that if $11!$ contains $a$ factors of $2$, then $x$ will contain $22-a$ factors of $2$.


Thus, once simplified, the expression will have no factors of $2$. It will also have no factors of $3$.


There is only one answer which is not even, which is $\boxed{\textbf{(D) }1925}$.

See Also

2013 AMC 10A (Problems • Answer Key • Resources)
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 (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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