$4-$ $[$ Formulas for abbreviated multiplication (other) $]$

Twenty-five coins are distributed into piles as follows. First, they are arbitrarily divided into two groups. Then any of the existing groups is again divided into two groups, and so on until each group consists of one coin. With each division of any group into two, the product of the number of coins in the two resulting groups is recorded. What can the sum of all recorded numbers be?
The first method. Let's represent the coins as points and connect each pair of points with a segment. We will get 25(25 - 1) : 2 $=300$ segments. Each time we divide one group of coins into two, we will erase all segments connecting points corresponding to coins that ended up in different groups. Suppose at some step we divide the coins of one of the existing groups into two groups of $x$ and $y$ coins. Then we erase $xy$ segments. This number is what we write down. Thus, the sum of the recorded numbers is the total number of erased segments. Since there were initially 300 segments, and in the end, all segments are erased, the total number of erased segments is 300.

The second method. Consider the variable quantity $S$, equal at each moment to half the sum of the squares of the number of coins in the piles. Initially,

$S=25^{2}: 2=312.5$, and at the very end $S=\left(1^{2}+\ldots+1^{2}\right): 2=12.5$. If a pile that had $x+y$ coins is divided into two piles of $x$ and $y$ coins, then $S$ decreases by $1 / 2(x+y)^{2}-1 / 2\left(x^{2}+y^{2}\right)$. Thus, with each division, the quantity $S$ decreases by the next recorded number. Therefore, the sum of all recorded numbers is equal to the total decrease in the quantity $S$, which is 312.5 - $12.5=300$.

## Answer

300.