Dividing the numbers 1200, 1640, and 1960 by the same (integer) number, we got remainders of 3, 2, and 7, respectively. What could the divisor have been?
Let the sought divisor be denoted by $d$. By subtracting the appropriate remainders from the dividends, we always get a number divisible by $d$; thus, $d$ can only be a common divisor of the numbers

$$
\begin{aligned}
1200-3=1197= & 3^{2} \cdot 7 \cdot 19, \quad 1640-2=1638=2 \cdot 3^{2} \cdot 7 \cdot 13 \quad \text { and } \\
& 1960-7=1953=3^{2} \cdot 7 \cdot 31
\end{aligned}
$$

and it must be such that the remainders 2, 3, 7 can occur. The greatest common divisor of these numbers is $3^{2} \cdot 7=63$, and all common divisors are the divisors of 63: $1,3,7,9,21,63$, so the sought divisor could only have been 9, 21, or 63.

László Hajnal (Bonyhád, Petőfi S. Gymn., IL. o. t.)