[ Division with remainder ]
[ Evaluation + example ]
Author: Polish A.
Chichikov is playing with Nozdryov. First, Nozdryov distributes 222 nuts into two small boxes. After looking at
the distribution, Chichikov names any integer \( N \) from 1 to 222. Then Nozdryov must, if necessary, move one or several nuts to an empty third box and show Chichikov one or two boxes, where the total is exactly \( N \) nuts. As a result, Chichikov will receive as many dead souls as the number of nuts Nozdryov moved.
What is the maximum number of souls Chichikov can guarantee for himself, regardless of how Nozdryov plays.
Upper bound. By placing 74 and 148 nuts in the boxes, Nozdryov can, for any $N$, move no more than 37. Indeed, $N$ can be written in the form $74 k+r$, where $k=0,1,2,3$, and $-37 \leq r0$, and the numbers 74, 148, and 222 can be collected with one or two boxes, without moving anything. If $r0,74 k$ $=0,74$ or 148. By taking a suitable box or not taking any, and adding a third box, where $r$ nuts are moved from the unselected one.

Lower bound. We will show that for any arrangement, there is an $N$ that will require moving no fewer than 37 nuts. Let there be $p$ and $q$ nuts in the boxes, $q \geq p$. If $p \geq 74$, then for $N=37$ it will be necessary to move no fewer than 37 nuts. If $p148$. To get $N=111$, one would either need to add more than 37 to $p$ nuts, or subtract more than 37 from $q$ nuts.

## Answer

37 souls.