8.3. Initially, the board had $n$ numbers written on it: $1,2, \ldots, n$. It is allowed to erase any two numbers on the board and write down the absolute value of their difference instead. What is the smallest number that can end up on the board after ( $n-1$ ) such operations a) for $n=111$; b) for $n=110$?
Answer: a) 0; b) 1. Solution. a) Using the following 55 operations, we can obtain 56 ones: $\{1,3-2,5-4, \ldots, 111-110\}=\{1,1, \ldots, 1\}$. From these, using 28 operations, we can get 28 zeros: $\{1-1,1-1, \ldots, 1-1\}=\{0,0, \ldots, 0\}$, and then after 27 operations - one 0. b) Note that with any operation, the parity of the sum of the numbers on the board does not change (since the numbers $a+b$ and $a-b$ have the same parity). Initially, the sum $1+2+\ldots+110$ was odd (it contains 55 odd addends and 55 even addends). Therefore, it is impossible to obtain 0 as a result of all operations. We can obtain one using a method similar to the one indicated in part a): specifically, $\{2-1,4-3, \ldots, 110-109\}=\{1,1, \ldots, 1\}$ - here we have 55 ones, and then $\{1,1-1,1-1, \ldots, 1-1\}=\{1,0, \ldots, 0\}$, i.e., we get one and 27 zeros, and then - one one.