## Problem 5.

On the board, 101 numbers are written: $1^{2}, 2^{2}, \ldots, 101^{2}$. In one move, it is allowed to erase any two numbers and write down the absolute value of their difference instead. What is the smallest number that can be obtained after 100 moves?

## Answer. 1
Solution. From four consecutive squares with 3 moves, the number 4 can be obtained:

$$
\begin{aligned}
& (n+1)^{2}-n^{2}=2 n+1 \\
& (n+3)^{2}-(n+2)^{2}=2 n+5 \\
& (2 n+5)-(2 n+1)=4
\end{aligned}
$$

In this way, we can get 24 quadruples from $6^{2}, 7^{2}, \ldots, 101^{2}$. From these, 20 quadruples can be removed by subtracting in pairs. From the numbers 4, 9, 16, 25, we can obtain

$$
14=(25-4)-(16-9)
$$

Finally, from the four quadruples, 1 and 14, we can obtain 1:

$4-(14-4-4-4)-1=1$.