The Apple Problem

Suppose there are $x$ red apples, $y$ green apples and $n$ stolen green apples, then $\begin{cases} 81 \leqslant x+y \leqslant 84 \\ x \geqslant 37 \\ n \geqslant 16 \end{cases}$ After $n$ green apples were stolen, there are less than 13 red apples that can't be matched as a pair (because there are no more green apples left for the rest of the red apples to match), from this we can conclude $y-n<x,\quad x-(y-n) \leqslant 12$ If $\frac{n}{4}$ red apples are switched to green, the number of unmatchable red apples should decrease by $\frac{n}{2}$ , we get $x-(y-n)-\frac{n}{2}=2$ $x-(y-n)=2+\frac{n}{2} \leqslant 12$ $16 \leqslant n \leqslant 20$ Rewrite $x-(y-n) \leqslant 12$ as $y-x\geqslant n-12\geqslant 4$ , add it with $y+x\geqslant 81$ , you can get $y \geqslant 42.5$ or simply $y \geqslant 43$ .

From $x-y\leqslant -4$ and $x+y\leqslant 84$ , we get $x\leqslant 40$ .

From $x+y\leqslant 84$ and $x\geqslant 37$ , we get $y\leqslant 47$ .

Thus, $\begin{cases} 37\leqslant x\leqslant 40 \\ 43\leqslant y \leqslant 47\end{cases}$

Since the number of red apples can be expressed as $4k+1$ , then $x-1$ must be a multiple of 4, therefore, $x$ can only be 37.

Meanwhile, from the problem we can know that $n$ is a multiple of 4, so the only possibilities for $n$ is 16 or 20. Substitute $n$ and $x$ it in the equation $x-(y-n)-\frac{n}{2}=2$ you will get $y=43,\,45$ Since $x+y\geqslant 81$ , thus $y=45$ , $n=20$ .

Therefore, the original number of apples in the box is $x+y=82$ .