Difference between revisions of "2013 AMC 12A Problems/Problem 17"
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Only one of the choices given is odd, <math>\boxed{\textbf{(D) }1925}</math>. | Only one of the choices given is odd, <math>\boxed{\textbf{(D) }1925}</math>. | ||
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== 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 00:37, 26 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. Here is why:
Consider the highest power of 2 that divides the starting number of coins, and consider how this value changes as each pirate takes their share. At each step, the size of the pile is multiplied by some 
. This means the highest power of 2 that divides the number of coins is continually decreasing (except once, briefly, when we multiply by 
for pirate 4, but it immediately drops again in the next step).
Therefore, if the 12th pirate's coin total were even, then it can't be the smallest possible value, because we can safely cut the initial pot (and all the intermediate totals) in half. We could continue halving the result until the 12th pirate's total is finally odd.
Only one of the choices given is odd, .
Note that if there had been multiple odd answers, we could repeat the reasoning above to see that the value also must not be divisible by 3.
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. 
