A cube with a side of 10 is divided into 1000 smaller cubes with an edge of 1. In each small cube, a number is written, and the sum of the numbers in each column of 10 cubes (in any of the three directions) is 0. In one of the cubes (denoted as $A$), the number 1 is written. Three layers, parallel to the faces of the cube (each layer having a thickness of 1), pass through cube A. Find the sum of all the numbers in the cubes that do not lie in these layers.
Through the given cube $A$ pass one horizontal layer $G$ and two vertical layers. The sum of all numbers in 81 vertical columns not included in the last two layers is 0. From this sum, we need to subtract the sum $S$ of the numbers lying in the cubes at the intersection of these columns with $G$ (there are 81 such cubes). These cubes are completely covered by nine columns lying in $G$. The sum of all numbers in these columns (which is 0) exceeds $S$ by the sum of nine numbers lying in a perpendicular column containing $A$. The latter sum is obviously -1. Hence, $S=1$. Finally, we have: $0-1=-1$.

## Answer

$-1$.