Difference between revisions of "2013 AMC 10A Problems/Problem 21"
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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>. | ||
| − | ==Solution 2== | + | ==Solution 2 (Using the answer choices)== |
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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> \frac{x \cdot 11!}{12^{11}}</math>. | ||
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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>. | ||
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There is only one answer which is not even, which is <math>\boxed{\textbf{(D) }1925}</math>. | There is only one answer which is not even, which is <math>\boxed{\textbf{(D) }1925}</math>. | ||
Revision as of 16:31, 7 August 2017
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 (Using the answer choices)
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 this expression, we must take out factors of and , since 
. 
has neither 
factors of , nor 
factors of . That means will substitute those factors, since the expression must have an integer value. Notice now how the numerator will have no factors of or .
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
