[ Arithmetic progression $]$  [ Extreme principle (other) . ]  Author: Shapovalov $A . B$.  In a $29 \times 29$ table, the numbers $1,2,3, \ldots, 29$ were written, each 29 times. It turned out that the sum of the numbers above the main diagonal is three times the sum of the numbers below this diagonal. Find the number written in the central cell of the table.
Above (below) the diagonal there are $29 \cdot 14=406$ numbers. It is not difficult to check that the sum of the 406 largest numbers in the table (16, 17, ..., 29, taken 29 times each) is exactly three times the sum of the 406 smallest numbers (1, 2, ..., 14, taken 29 times each). Therefore, all the numbers on the diagonal are 15.

## Answer

15. 

## Problem