Fomin D:

The numbers $1, 1/2, 1/3, \ldots, 1/100$ are written on the board. 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 an allowable 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 equal to $(1+1 / 1)\left(1+\frac{1}{2}\right) \ldots(1+1 / 100)-1=2 / 1 \cdot 3 / 2 \cdot \ldots \cdot 101 / 100-1=$ $101-1=100$.

## Answer

100. 

Author: Shapovalov A.V.

In the Simpleton Primary School, there are only 20 children. Any two of them have a common grandfather. Prove that one of the grandfathers has at least 14 grandchildren in this school.

## Solution

Consider any student $A$ and his two grandfathers $X, Y$. If not all students are grandchildren of one of them (in which case there is nothing to prove), then there is a student $B$ who is not a grandson of $X$ (then he is necessarily a grandson of $Y$), and a student $C$ who is not a grandson of $Y$ (hence, a grandson of $X$). Students $B$ and $C$ have a common grandfather $Z$.

There are no other grandfathers besides $X, Y$, and $Z$ for the students: a grandson of such a grandfather would not have a common grandfather with one of the three students $A, B, C$. Twenty grandchildren are distributed among three grandfathers, so (counting two grandfathers per grandchild) at least one of them has no less than $2 / 3 \cdot 20$, that is, no less than 14 grandchildren.

Send a comment