On each of a hundred cards, a number different from zero is written, such that each number is equal to the square of the sum of all the others.

What are these numbers

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Each of these numbers is the square of a number different from zero, so all the written numbers are positive. Let's prove that they are all the same.

The first way. Suppose that the number written on one of the cards is greater than the number written on another. Set aside the card with the larger number. By condition, this number is equal to the square of the sum of the other numbers. Swap the cards with the larger and smaller numbers. Then the set-aside number has decreased, while the sum of all other numbers (and thus its square) has increased, and the equality can no longer hold.

The second way. Let the numbers written be $a_{1}, a_{2}, \ldots, a_{100}$. We will prove, for example, that $a_{1}=a_{2}$. $a_{2}-a_{1}=\left(a_{1}+a_{3}+\ldots+a_{100}\right)^{2}-\left(a_{2}+a_{3}+\ldots+a_{100}\right)^{2}=\left(a_{1}+a_{2}+2 a_{3}+\ldots+2 a_{100}\right)\left(a_{1}-a_{2}\right)$. The sum in the left parenthesis is positive, so the numbers $a_{2}-a_{1}$ and $a_{1}-a_{2}$ are either both zero or have the same sign. The latter is impossible.

Let the number written on each card be denoted by $x$. Then $(99 x)^{2}=x$. Since $x \neq 0$, then $x=1 / 99^{2}$.

## Answer

One hundred numbers, each equal to $1 / 99^{2}$.