Difference between revisions of "2013 AMC 12A Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | + | ===Solution 1=== | |
The first pirate takes <math>\frac{1}{12}</math> of the <math>x</math> coins, leaving <math>\frac{11}{12} x</math>. | The first pirate takes <math>\frac{1}{12}</math> of the <math>x</math> coins, leaving <math>\frac{11}{12} x</math>. | ||
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We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, <math>x</math> is the denominator, leaving <math>1925</math> coins for the twelfth pirate. | We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, <math>x</math> is the denominator, leaving <math>1925</math> coins for the twelfth pirate. | ||
| + | |||
| + | ===Solution 2=== | ||
| + | The answer cannot be an even number. | ||
| + | |||
| + | Consider the prime factorization of the starting number of coins. This number will be repeatedly multiplied by <math>\frac{n}{12}</math>. At every step, we are only removing twos from the prime factorization, never adding them (except in a single case, when we multiply by <math>\frac{2}{3}</math> for pirate 4, but that 2 is immediately removed again in the next step). | ||
| + | |||
| + | Therefore, if the 12th pirate's coin total were even, this can't be the smallest possible value; we can cut the initial pot in half and safely cut all the intermediate totals in half. So this number must be odd. | ||
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| + | Only one of the choices given is odd, <math>1925</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:19, 25 January 2018
Problem 17
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
Solution 1
The first pirate takes 
of the 
coins, leaving 
.
The second pirate takes 
of the remaining coins, leaving 
.
Note that
All the 2s and 3s cancel out of 
, leaving
in the numerator.
We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, is the denominator, leaving 
coins for the twelfth pirate.
Solution 2
The answer cannot be an even number.
Consider the prime factorization of the starting number of coins. This number will be repeatedly multiplied by 
. At every step, we are only removing twos from the prime factorization, never adding them (except in a single case, when we multiply by 
for pirate 4, but that 2 is immediately removed again in the next step).
Therefore, if the 12th pirate's coin total were even, this can't be the smallest possible value; we can cut the initial pot in half and safely cut all the intermediate totals in half. So this number must be odd.
Only one of the choices given is odd, .
See also
| 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. 
