## Subject 1.

If we divide the numbers 3625 and 1611 by the number $\overline{a b c}$, we get the same remainder. Determine the number $\overline{a b c}$.
## Subject 1.

$$
\begin{array}{ll}
3625=\overline{a b c} \cdot x+r \text { and } 1611=\overline{a b c} \cdot y+r, \quad r<\overline{a b c} & \ldots \text { 2p. } \\
\text { By subtracting the two relations we get } 2014=\overline{a b c} \cdot(x-y) & \ldots \text { 1p. } \\
\text { Therefore } \overline{a b c} \text { is a divisor of } 2014 & \ldots \text { 1p. } \\
2014=2 \cdot 19 \cdot 53 \text {, so } \overline{a b c}=106 & \ldots \text { 3p. }
\end{array}
$$