Kanel-Belov A.Y.

A cube with a side of 20 is divided into 8000 unit cubes, and a number is written in each cube. It is known that in each column of 20 cubes, parallel to the edge of the cube, the sum of the numbers is 1 (columns in all three directions are considered). In some cube, the number 10 is written. Through this cube, three layers $1 \times 20 \times 20$, parallel to the faces of the cube, pass. Find the sum of all numbers outside these layers.
Through the given cube K, one horizontal layer G and two vertical layers pass. The sum of all numbers in 361 vertical columns, not included in the last two layers, is 361. 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 361 such cubes). These cubes are completely covered by 19 columns lying in G. The sum of all numbers in these columns (which is 19) exceeds $S$ by the sum of 19 numbers lying in a perpendicular column containing K. The latter sum is clearly $1-10=-9$. Therefore, $S=19-(-9)=28$. Finally, we have: $361-28=333$.

## Answer

333.