Fomin D.

On the board, the numbers $1, 1/2, 1/3, \ldots, 1/100$ are written. We choose two arbitrary numbers $a$ and $b$ from those written on the board, erase them, and write the number

$a + b + ab$. We perform this operation 99 times until only one number remains. What is this number? Find it and prove that it does not depend on the sequence of number selections.
If $a_{1}, a_{2}, \ldots, a_{n}$ are numbers written on the board, then the value $\left(1+a_{1}\right)\left(1+a_{2}\right) \ldots\left(1+a_{n}\right)$ does not change under a permissible operation. Indeed, if $a$ and $b$ are replaced by $a+b+a b$, then the factors not containing $a$ and $b$ do not change, and the product $(1+a)(1+b)$ is replaced by the equal number

$1+a+b+a b$. Therefore, the last number on the board is $\left(1+\frac{1}{1}\right)(1+1 / 2) \ldots\left(1+\frac{1}{100}\right)-1=2 / 1 \cdot 3 / 2 \cdot \ldots \cdot 101 / 100-1=$ $101-1=100$.

## Answer

100.