Difference between revisions of "2015 AMC 8 Problems/Problem 20"

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</math>
 
</math>
  
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==Solution==
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So let there be <math>x</math> pairs of <math>1</math> dollar socks, <math>y</math> pairs of <math>3</math> dollar socks, <math>z</math> pairs of <math>4</math> dollar socks.
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We have <math>x+y+z=12</math>, <math>x+3y+4z=24</math>, and <math>x,y,z \ge 1</math>.
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Now we subtract to find <math>2y+3z=12</math>, and <math>y,z \ge 1</math>.
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It follows that <math>y</math> is a multiple of <math>3</math> and <math>2y</math> is a multiple of <math>6</math>, so since <math>0<2y<12</math>, we must have <math>2y=6</math>.
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Therefore, <math>y=3</math>, and it follows that <math>z=2</math>. Now <math>x=12-y-z=12-3-2=\boxed{\textbf{(D)}~7}</math>, as desired.
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2015|num-b=19|num-a=21}}
 
{{AMC8 box|year=2015|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:41, 25 November 2015

Ralph went to the store and bought 12 pairs of socks for a total of $24. Some of the socks he bought cost $1 a pair, some of the socks he bought cost $3 a pair, and some of the socks he bought cost $4 a pair. If he bought at least one pair of each type, how many pairs of $1 socks did Ralph buy?

$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution

So let there be $x$ pairs of $1$ dollar socks, $y$ pairs of $3$ dollar socks, $z$ pairs of $4$ dollar socks.

We have $x+y+z=12$, $x+3y+4z=24$, and $x,y,z \ge 1$.

Now we subtract to find $2y+3z=12$, and $y,z \ge 1$. It follows that $y$ is a multiple of $3$ and $2y$ is a multiple of $6$, so since $0<2y<12$, we must have $2y=6$.

Therefore, $y=3$, and it follows that $z=2$. Now $x=12-y-z=12-3-2=\boxed{\textbf{(D)}~7}$, as desired.

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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 AJHSME/AMC 8 Problems and Solutions

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