Problem 9.6. The teacher wrote a number on the board. Sasha decided to divide it by 102 with a remainder, and Masha - by 103. It turned out that the quotient obtained by Sasha and the remainder obtained by Masha add up to 20. What remainder did Sasha get? List all possible options.
Answer: 20.

Solution. Let $n$ be the number on the board, divide it by 102 and by 103 with remainders:

$$
n=102 a+b ; \quad n=103 c+(20-a)
$$

Subtract the second equation from the first:

$$
0=103(a-c)+(b-20)
$$

From this, it follows that $b-20$ is divisible by 103. Since $0 \leqslant b \leqslant 101$, we get that $b=20$.

Notice also that, for example, for the number $n=122$, the remainder $b=20$ is indeed possible.