The integers from 1 to 20 are written on the board. We can erase any two numbers, $a$ and $b$, and write the number $a+b-1$ in their place. What number will be written on the board after 19 operations? And if we wrote $a+b+ab$ instead of $a+b-1$?

## Extremum Principle

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At any time, if the integers $n_{1}, \ldots, n_{k}$ are written on the board, we consider the sum $S=n_{1}+n_{2}+\ldots+n_{k}-k$. Then, with each operation, the sum $S$ does not change. In particular, after 19 operations, there will be a unique integer $n$ such that $n=S+1=(1+2+\ldots+20-20)+1=\frac{19 \times 20}{2}+1=191$.

If we write $a+b+a b$ instead of $a+b-1$, the interesting quantity to observe is no longer $S$, but $P=\left(n_{1}+1\right)\left(n_{2}+1\right) \cdots\left(n_{k}+1\right)$. Indeed, with each operation, we divide $P$ by $(a+1)(b+1)$ before multiplying it by $a+b+a b+1=(a+1)(b+1)$, so that $P$ does not change. Consequently, after 19 operations, there will be a unique integer $n$ such that $n=P-1=2 \times 3 \times \cdots \times 21-1=21!-1$.