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

Revision as of 16:34, 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 $2$ . This also holds for factors of $3$ .

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

Looking at the answer choices, there is only one answer which is not even, which is $\boxed{\textbf{(D) }1925}$ .

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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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

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

Revision as of 16:33, 7 August 2017 (view source) Wannabecharmander (talk \| contribs) (→‎Solution 2 (Using the answer choices)) ← Older edit		Revision as of 16:34, 7 August 2017 (view source) Wannabecharmander (talk \| contribs) (→‎Solution 2 (Using the answer choices)) Newer edit →
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−	~~There~~ is only one answer which is not even, which is <math>\boxed{\textbf{(D) }1925}</math>.	+	Looking at the answer choices, there is only one answer which is not even, which is <math>\boxed{\textbf{(D) }1925}</math>.

	==See Also==		==See Also==

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

Revision as of 16:34, 7 August 2017

Contents

Problem

Solution 1

Solution 2 (Using the answer choices)

See Also