Difference between revisions of "2013 AMC 10A Problems/Problem 21"
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Revision as of 06:49, 4 October 2020
Contents
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 . This means that if contains factors of , then will contain factors of . This also holds for factors of .
Thus, once simplified, the expression will have no factors of . It will also have no factors of .
Looking at the answer choices, there is only one answer which is not even, which is .
Solution 3
We know that the 11th pirate takes of what is left from the 10th pirate, so we have the 12th pirate taking of what is left from the 10th pirate. Similarly, since the 10th pirate takes of what is left from the 9th pirate, we have the 11th pirate taking of what is left from the 9th pirate. Thus, the 12th pirate takes of what is left from the 9th pirate. Repeating the method, we can find that the 12th pirate takes of what is left from the 1st pirate, or of the total amount of coins.
Now canceling out the denominator with the numerator as possible, we are left with with some factors of 12 in the denominator. For this fraction to be an integer, the smallest possible number of coins is the same as the denominator, so the numerator is the number of coins taken by the 12th pirate, or
~ Nafer
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=-3d31c-7xm8
~dolphin7
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.