A school with $100$ students offers French and Spanish as its language courses. Twice as many students are in the French class as the Spanish class. Three times as many students are in both classes as are in neither class. The number of students in both classes is even, and fewer than $10$ students are in neither class. How many students are taking Spanish?
Let the number of students taking French be $f$, the number of those taking Spanish be $s$, the number of students taking both be $x$, and the number of students taking neither be $n$. Then we have $100 = f+s-x+n$ (by PIE). We are also given that $f=2s$ and $x=3n$, where $n<10$.
Substituting some values, $100 = 3s-2n$ or alternatively, $100+2n = 3s$. Since $n<10$, we can see which values work: $n=1, 4$ or $7$.
When $n=1$, $102=3s$ and thus $s=34$, and $f=68$. When $n=4$, $108=3x$ and thus $s=36$ and $f=72$. When $n=7$, $114=3s$ and thus $s=38$ and $f=76$. All of these situations satisfy the condition that $f\equiv 0\mod 2$ & $s\equiv 0\mod 2$, and satisfy that $x=3n$ (when calculated using PIE) as well. However, it seems that there should only be one set of $(f,s)$ values that should work. What am I missing or what have I done wrong?
As already discussed in the comments, the problem statement is ambiguous. It can be read to mean that the number of students in the French class is even and the number of students in the Spanish class is even, but from the fact that this leads to more than one solution, we can assume that the intended meaning is that the number of students who visit both classes is even. Then the solution is unique.