Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 10A Problems/Problem 21"

(→‎Solution)
(→‎Solution)
Line 6: Line 6:
 
<math> \textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850 </math>
 
<math> \textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850 </math>
  
−
==Solution==
+
==Solution 1==
  
 
Let <math>x</math> be the number of coins.  After the <math>k^{\text{th}}</math> pirate takes his share, <math>\frac{12-k}{12}</math> of the original amount is left.  Thus, we know that
 
Let <math>x</math> be the number of coins.  After the <math>k^{\text{th}}</math> pirate takes his share, <math>\frac{12-k}{12}</math> of the original amount is left.  Thus, we know that
Line 16: Line 16:
  
 
Thus, we know the denominator is canceled out, so the number of gold coins received is going to be the product of the numerators, <math>11 \cdot 5 \cdot 7 \cdot 5 = \boxed{\textbf{(D) }1925}</math>.
 
Thus, we know the denominator is canceled out, so the number of gold coins received is going to be the product of the numerators, <math>11 \cdot 5 \cdot 7 \cdot 5 = \boxed{\textbf{(D) }1925}</math>.
 +
 +
==Solution 2==
 +
 +
Solution <math>1</math> mentioned the expression <math>x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot ... \cdot \frac{1}{12}</math>. Note that this is equivalent to <math>x \cdot \frac{11!}{12^{11}}</math>. We can compute the amount of factors of <math>2</math>, <math>3</math>, <math>5</math>, etc. but this is not necessary. To minimize
 +
 +
There is only one answer which is not even, which is <math>\boxed{\textbf{(D) }1925}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 17:26, 7 August 2017

Problem

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 1

Let $x$ be the number of coins. After the $k^{\text{th}}$ pirate takes his share, $\frac{12-k}{12}$ of the original amount is left. Thus, we know that

$x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot \frac{9}{12} \cdot \frac{8}{12} \cdot \frac{7}{12} \cdot \frac{6}{12} \cdot \frac{5}{12} \cdot \frac{4}{12} \cdot \frac{3}{12} \cdot \frac{2}{12} \cdot \frac{1}{12}$ must be an integer. Simplifying, we get


$x \cdot \frac{11}{12} \cdot \frac{5}{6} \cdot \frac{1}{2} \cdot \frac{7}{12} \cdot \frac{1}{2} \cdot \frac{5}{12} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{6} \cdot \frac{1}{12}$. Now, the minimal $x$ 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 $12^{\text{th}}$ pirate receives, as he receives $\frac{12}{12} = 1 =$ 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, $11 \cdot 5 \cdot 7 \cdot 5 = \boxed{\textbf{(D) }1925}$.

Solution 2

Solution $1$ mentioned the expression $x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot ... \cdot \frac{1}{12}$. Note that this is equivalent to $x \cdot \frac{11!}{12^{11}}$. We can compute the amount of factors of $2$, $3$, $5$, etc. but this is not necessary. To minimize

There is only one answer which is not even, which is $\boxed{\textbf{(D) }1925}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_21&oldid=86932"