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

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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.
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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>\boxed{\textbf{(D) }1925}</math> coins for the twelfth pirate.
  
 
===Solution 2===
 
===Solution 2===

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.

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, $1925$.

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

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