4. On the board, the numbers $1 / 1, 1 / 2, 1 / 3, 1 / 4, \ldots 1 / 100$ are written.

It is allowed to erase any two numbers a, b and write $a b + a + b$ in their place,

then do the same with any two of the remaining numbers, and so on. What number can be left last?
Solution. If initially on the board there were not one hundred, but three numbers: a, b, c, then at the end on the board there would be the number abc + ab + bc + ac + a + b + c. That is, the sum of all possible monomials composed of the numbers a, b, c, taken no more than once each.

In the case when one hundred numbers are written on the board. It is easy to see that the last number on the board will be a number represented as the sum of (2^100)-1 different monomials of the form (1/a_1)*(1/a_2)*....(1/a_k) for all possible natural numbers $\mathrm{k}$ in $[1,100]$ and a_i in [1,100], with a_i < a_i+1 for all i. The schoolchildren will need to prove this.

Now notice that this sum can be represented as $(1+1 / 1) *(1+1 / 2) *(1+1 / 3) * \ldots *(1+1 / 100)-1$.

Transform it: $(1+1 / 1) *(1+1 / 2) *(1+1 / 3) * . . .(1+1 / 100)-1=$ $(2 / 1) *(3 / 2) *(3 / 4) * \ldots *(101 / 100)-1=101-1=100$.

## Criteria

+/2 or -+ - it is proven that the last remaining number is representable as the correct sum of monomials. +- - it is noted that this sum is equal to $(1+1 / 1) *(1+1 / 2) *(1+1 / 3)^{*} . . *(1+1 / 100)$ -