Find the smallest positive integer that, when multiplied by 1999, the last four digits of the resulting number are 2001.
Let $n$ be a positive integer such that $1999 n$ ends in 2001. Then their difference

$$
1999 n-2001=\ldots 0000
$$

ends in four zeros, and thus is divisible by - among others - 2000.

Transform the difference (1) as follows:

$$
1999 n-2001=2000(n-1)-(n+1)
$$

Since the left side is divisible by 2000, the right side must also be divisible, that is,

$$
2000 \mid n+1
$$

The possible values of $n$ are: 1999, 3999, 5999, ... When these are multiplied by 1999, $5999 \cdot 1999=11992001$ is the first one that ends in 2001, so 5999 is the smallest $n$.

Márton Hablicsek (Fazekas M. Főv. Gyak. Gimn., 8th grade)