Difference between revisions of "2013 AMC 10A Problems/Problem 21"
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==Solution== | ==Solution== | ||
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| + | 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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| + | <math>x * \frac{11}{12} * \frac{10}{12} * \frac{9}{12} * \frac{8}{12} * \frac{7}{12} * \frac{6}{12} * \frac{5}{12} * \frac{4}{12} * \frac{3}{12} * \frac{2}{12} * \frac{1}{12}</math> must be an integer. Simplifying, we get | ||
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| + | <math>x * \frac{11}{12} * \frac{5}{6} * \frac{1}{2} * \frac{7}{12} * \frac{1}{2} * \frac{5}{12} * \frac{1}{3} * \frac{1}{4} * \frac{1}{6} * \frac{1}{12}</math>. 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 <math>12^{\text{th}}</math> pirate receives, as he receives <math>\frac{12}{12} = 1 =</math> all of what is remaining. | ||
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| + | 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 * 5 * 7 * 5</math> = <math>1925</math>, <math>\textbf({D})</math>. | ||
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==See Also== | ==See Also== | ||
Revision as of 22:05, 7 February 2013
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
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 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 pirate receives, as he receives 
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, 
= 
, 
.
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 | |
