Problem 206

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