The numbers $1, 1/2, \ldots, 1/n$ are written on a board. It is allowed to erase any two numbers $a$ and $b$ and replace them with the number $ab + a + b$.

What number will remain after $n-1$ such operations?
During this operation, the product of the numbers increased by 1 is preserved.

## Solution

Let at some point the numbers $a, b, \ldots, z$ be written on the board. Consider the product $P=(a+1)(b+1) \ldots(z+1)$. After replacing the pair of numbers $a, b$ with $ab+a+b$, the term $(a+1)(b+1)$ in the product will be replaced by $ab+a+b+1=(a+1)(b+1)$. Thus, the operation does not change the value of $P$. Therefore, the number that remains at the end does not depend on the order of the operations and is equal to

$(1+1)(1/2+1) \ldots(1/n+1)-1=2/1 \cdot 3/2 \cdot \ldots \cdot (n+1)/n-1=(n+1)-1=n$.

## Answer

n.