Problem of the Week
Problem C and Solution
Will it be Four or Seven?

Problem

Sophia has been making two types of necklaces: small necklaces that contain four beads each and large necklaces that contain seven beads each.

After creating a certain number of small and large necklaces, a total of $99$ beads have been used. Determine all possibilities for how many of each type of necklace Sophia has made.

Solution

Let $S$ represent the number of small necklaces and $L$ represent the number of large necklaces she has made. Since $S$ and $L$ represent numbers of necklaces, both must be integers greater than or equal to $0.$ Since the small necklaces use $4$ beads each, $S$ necklaces would use $4\times S$ or $4S$ beads in total. Since the large necklaces use $7$ beads each, $L$ necklaces would use $7\times L$ or $7L$ beads in total. Since a total of $99$ beads have been used, $4S+7L=99.$

We can also determine a maximum value for $L$. Since each large necklace uses $7$ beads, $99÷7\approx 14.1$, and $L$ must be an integer, we know that $L$ must be less than or equal to $14.$ Thus, $L$ is an integer greater than or equal to $0$ and less than or equal to $14$. We could at this point check all of the possible integer values for $L$ from $0$ to $14.$ However, we can narrow down the possibilities even more.

In the equation, $4S+7L=99$, $4S$ will always be an even integer since $4$ times any integer is always even. We have the even integer $4S$ plus $7L$ is equal to the odd integer $99$. This means that $7L$ must be an odd integer. (The sum of an even integer and an even integer is an even integer, not an odd integer.) For $7L$ to be an odd integer, $L$ must be odd. (If $L$ is even, $7L$ would be even.) This observation reduces the possible values for $L$ to the odd positive integers between $0$ and $14$, namely $1$, $3$, $5$, $7$, $9$, $11$, $13$. For each possible value of $L$, we now determine $7L$, the total number of large beads used, $4S=99-7L$, the total number of small beads used, and finally $S=(99-7L)÷4$, the number of small necklaces for that value of $L.$ If $S$ is a non-negative integer, then we have found a valid possibility.

Therefore, Sophia has either made $1$ large necklace and $23$ small necklaces, or $5$ large necklaces and $16$ small necklaces, or $9$ large necklaces and $9$ small necklaces, or $13$ large necklaces and $2$ small necklaces.