Bakayev E.V.

About the group of five people, it is known that:

Alyosha is 1 year older than Alekseev,

Borya is 2 years older than Borisov,

Vasya is 3 years older than Vasiliev,

Grisha is 4 years older than Grigoryev, and there is also Dima and Dmitriev in this group.

Who is older and by how many years: Dima or Dmitriev?
The sum of the ages of Alyosha, Borya, Vasya, Grisha, and Dima is equal to the sum of the ages of Alexeev, Borisov, Vasilyev, Grigoryev, and Dmitriev. Therefore, Dmitriev is older than Dima by \(1+2+3+4=10\) years.

## Answer

\[
\begin{aligned} &\({[\quad\) The Pigeonhole Principle (other). \(] } \\ &\) Problem \(\underline{32787}\) Topics: \\ &\({[\underline{\text { Pairings and groupings; bijections }}] } \\ & {[\) Decimal number system \(] }\end{aligned}
\]

Can 57 different two-digit numbers be found such that the sum of any two of them is not equal to 100?

## Solution

Suppose such 57 two-digit numbers can be chosen. Then from each pair of the form (n, 100-n), where n takes values from 10 to 49, no more than one number is chosen; there are 40 such pairs, so no more than 41 numbers are chosen from the numbers 10 to 90 (the number 50 could also be chosen). There are still 9 numbers from 91 to 99 left; thus, in total, no more than 50 numbers are chosen, but we assumed that 57 numbers were chosen, which is a contradiction. Therefore, our assumption was incorrect.

## Answer

No, it is not possible.