1. On the board, the numbers $1,2,3, \ldots, 19$ are written. The following procedure is performed: any two numbers $x$ and $y$ are erased and the number $x+y-1$ is written in their place. The procedure continues until only one number remains. What is that number?
Solution. Let $a_{k}$ denote the difference between the sum of all numbers on the board after the $k$-th step, and the number of numbers written on the board after the $k$-th step. For the initial position (zeroth step), we have:

$$
a_{0}=(1+2+3+4+\ldots+19)-19=171
$$

We will prove that $a_{k}=a_{p}$ for any $k$ and $p$ ($k \leq 18$ and $p \leq 18$). Suppose that in the $(k+1)$-th step, the numbers $x$ and $y$ are written, and in their place, the number $x+y-1$ is written. Then we get:

$$
a_{k}=x+y+\sum z_{i}-(19-k)
$$

and

$$
a_{k+1}=(x+y-1)+\sum z_{i}-(19-k-1)=x+y+\sum z_{i}-(19-k)
$$

Thus: $a_{k}=a_{p}$ for any $k$ and $p$ ($k \leq 18$ and $p \leq 18$). According to this,

$$
a_{18}=d-1=a_{0}=171
$$

Finally, the number 172 is obtained.