# Problem 6.

Find the smallest natural number $n$, for which the number $n+2018$ is divisible by 2020, and the number $n+2020$ is divisible by 2018.
Answer: 2034142

Solution. By the condition $n+2018=2020 m, n+2020=2018 k$, hence $1009 k-1010 m=1$. The solution to this Diophantine equation is: $k=-1+1010 p, m=-1+1009 p$. Therefore, $n+2018=$ $2020(-1+1009 p)=-2020+2020 \cdot 1009 p$, from which $n=-2018-2020+2020 \cdot 1009 p$. The smallest natural $n$ is $n=2020 \cdot 1009-2020-2018=2034142$.

## B-2

Find the smallest natural number $n$ for which the number $n+2019$ is divisible by 2020, and the number $n+2020$ is divisible by 2019.

Answer: 4074341

## B-3

Find the smallest natural number $n$ for which the number $n+2019$ is divisible by 2021, and the number $n+2021$ is divisible by 2019.

Answer: 4076359

## B-4

Find the smallest natural number $n$ for which the number $n+2020$ is divisible by 2022, and the number $n+2022$ is divisible by 2020.

Answer: 2038178

## Lomonosov Olympiad for Schoolchildren in Mathematics

Preliminary stage $2020 / 21$ academic year for $7-8$ grades

## B-1

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