Difference between revisions of "2013 AMC 10A Problems/Problem 21"
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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> | <math> \textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850 </math> | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>x</math> be the number of coins. After the <math>k^{\text{th}}</math> pirate takes his share, <math>\frac{12-k}{12}</math> of the original amount is left. Thus, we know that | Let <math>x</math> be the number of coins. After the <math>k^{\text{th}}</math> pirate takes his share, <math>\frac{12-k}{12}</math> of the original amount is left. Thus, we know that | ||
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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>. | 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>. | ||
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+ | ==Solution 2== | ||
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+ | Solution <math>1</math> mentioned the expression <math>x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot ... \cdot \frac{1}{12}</math>. Note that this is equivalent to <math>x \cdot \frac{11!}{12^{11}}</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 | ||
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+ | There is only one answer which is not even, which is <math>\boxed{\textbf{(D) }1925}</math>. | ||
==See Also== | ==See Also== |
Revision as of 16:26, 7 August 2017
Contents
[hide]Problem
A group of pirates agree to divide a treasure chest of gold coins among themselves as follows. The pirate to take a share takes 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 pirate receive?
Solution 1
Let be the number of coins. After the pirate takes his share, of the original amount is left. Thus, we know that
must be an integer. Simplifying, we get
. Now, the minimal 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 pirate receives, as he receives 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, .
Solution 2
Solution mentioned the expression . Note that this is equivalent to . We can compute the amount of factors of , , , etc. but this is not necessary. To minimize
There is only one answer which is not even, which is .
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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