Problem 1. Find the last three digits of the non-zero number $n$, knowing that by dividing $29 \cdot n$ by 250 we get a remainder of 67, and by dividing $23 \cdot n$ by 200 we get a remainder of 29.

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Solution:

From the division theorem with remainder, we have the relations:

$$
\begin{cases}29 n=250 x+67 \\ 23 n=200 y+29\end{cases}
$$

$1 p$

Multiplying relation (1) by 4 and relation (2) by 5, we get:

$$
\left\{\begin{array}{l}
116 n=1000 x+268 \\
115 n=1000 y+145
\end{array}\right.
$$

$2 p$

Subtracting relation (4) from relation (3) member by member, we get:

![](https://cdn.mathpix.com/cropped/2024_06_07_22cfa7b831b4c749126cg-1.jpg?height=76&width=1538&top_left_y=1506&top_left_x=246)

$U_{3}(n)=U_{3}(1000(x-y)+123)=\underline{123}$

$.2 p$