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2015 AMC 8 Problems/Problem 20

Problem

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?

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

Solutions

Solution 1

So, let there be $x$ pairs of $$1$ socks, $y$ pairs of $$3$ socks, and $z$ pairs of $$4$ 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 $2y$ is a multiple of $3$ and $3z$ is a multiple of $3$. Since sum of 2 multiples of 3 = multiple of 3, so 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.

Solution 2

Since the total cost of the socks was $$24$ and Ralph bought $12$ pairs, the average cost of each pair of socks is $\frac{$24}{12} = $2$.

There are two ways to make packages of socks that average to $$2$. You can have:

$\bullet$ Two $$1$ pairs and one $$4$ pair (package adds up to $$6$)

$\bullet$ One $$1$ pair and one $$3$ pair (package adds up to $$4$)

Now, we need to solve \[6a+4b=24,\] where $a$ is the number of $$6$ packages and $b$ is the number of $$4$ packages. We see our only solution (that has at least one of each pair of sock) is $a=2, b=3$, which yields the answer of $2\times2+3\times1 = \boxed{\textbf{(D)}~7}$.

Solution 3

Since there are 12 pairs of socks, and Ralph bought at least one pair of each, there are $12-3=9$ pairs of socks left. Also, the sum of the three pairs of socks is $1+3+4=8$. This means that there are $24-8=16$ dollars left. If there are only $1$ dollar socks left, then we would have $9\cdot1=9$ dollars wasted, which leaves $7$ more dollars. If we replace one pair with a $3$ dollar pair, then we would waste an additional $2$ dollars. If we replace one pair with a $4$ dollar pair, then we would waste an additional $3$ dollars. The only way $7$ can be represented as a sum of $2$s and $3$s is $2+2+3$. If we change $3$ pairs, we would have $6$ pairs left. Adding the one pair from previously, we have $\boxed{(\text{D})~7}$ pairs.

Solution 4

Let the amount of $1$ dollar socks be $a$, $3$ dollar socks be $b$, and $4$ dollar socks be $c$. We then know that $a+b+c=12$ and $a+3b+4c=24$. We can make $a+b+c=12$ into $a=12-b-c$ and then plug that into the other equation, producing $12-b-c+3b+4c=24$ which simplifies to $2b+3c=12$. It's not hard to see $b=3$ and $c=2$. Now that we know $b$ and $c$, we know that $a=7$, meaning the number of $1$ dollar socks Ralph bought is $\boxed{\textbf{(D)} 7}$.

Solution 5 (Guess and check)

If Ralph bought one sock of each kind, he already used $$8$, so there are $$16$ left and 9 socks. If we split the $$16$ into four $$4$ sections, (as it is the smallest possible number that 1, 3, 4, can make in different ways that in all use at least each of the numbers once,) if Ralph bought a $$3$ pair, he would need to buy a $$1$ pair in order for it to add up to a multiple of four. Similarly, if Ralph bought a $$1$ pair, he would either need to buy three $$1$ pairs or a $$3$ pair. If Ralph bought a $$4$ pair, it would already make a group. Now, the problem is just how we can split 9 into 4 groups of 1, 2, or 4. We clearly see that $1 + 2 + 2 + 4 = 9$, or a $$4$ pair, two $$3$ pairs, and six $$1$ pairs. Because we subtracted the necessary one of each kind, there are two $$4$ pairs, three $$3$ pairs, and seven $$1$ pairs. Therefore, the number of $$1$ pairs Ralph bought is $\boxed{\textbf{(D)}~7}$. ~strongstephen

Video Solution by Pi Academy (Fast and Easy)

https://youtu.be/ugbi97qGScY?si=t3zYt83JEwa0PXtJ

~ jj_empire

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/toZI27nNHeQ

~Education, the Study of Everything

Video Solution

https://youtu.be/hvnVuLbveJs

~savannahsolver

Video Solution

https://youtu.be/rQUwNC0gqdg?t=2187

~pi_is_3.14

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
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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