Difference between revisions of "2013 AMC 12A Problems/Problem 17"

(Solution 1)
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===Solution 2===
 
===Solution 2===
The answer cannot be an even number.  
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The answer cannot be an even number. Here is why:
  
 
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).
 
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).
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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.
 
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, <math>1925</math>.
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Only one of the choices given is odd, <math>\boxed{\textbf{(D) }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 14:22, 25 January 2018

Problem 17

A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ 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 $12^{\text{th}}$ pirate receive?

$\textbf{(A)} \ 720 \qquad  \textbf{(B)} \ 1296 \qquad  \textbf{(C)} \ 1728 \qquad  \textbf{(D)} \ 1925 \qquad  \textbf{(E)} \ 3850$

Solution

Solution 1

The first pirate takes $\frac{1}{12}$ of the $x$ coins, leaving $\frac{11}{12} x$.

The second pirate takes $\frac{2}{12}$ of the remaining coins, leaving $\frac{10}{12}*\frac{11}{12}*x$.

Note that

$12^{11} = (2^2 * 3)^{11} = 2^{22} * 3^{11}$

$11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2$


All the 2s and 3s cancel out of $11!$, leaving

$11 * 5 * 7 * 5 = 1925$

in the numerator.


We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, $x$ is the denominator, leaving $\boxed{\textbf{(D) }1925}$ coins for the twelfth pirate.

Solution 2

The answer cannot be an even number. Here is why:

Consider the prime factorization of the starting number of coins. This number will be repeatedly multiplied by $\frac{n}{12}$. At every step, we are only removing twos from the prime factorization, never adding them (except in a single case, when we multiply by $\frac{2}{3}$ 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, $\boxed{\textbf{(D) }1925}$.

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
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

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