3. Find the last three digits of the non-zero natural number $n$, knowing that when $29 n$ is divided by 250, the remainder is 67, and when $23 n$ is divided by 200, the remainder is 29.

(Gazeta Matematica)
Subject 3, author Gazeta Matematica

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

| Details of solution | Associated grading |
| :--- | :---: |
| $29 \mathrm{n}=250 \cdot \mathrm{K}+67$ and $23 \mathrm{n}=200 \cdot \mathrm{Q}+29$ | 1 |
| $29 \mathrm{n} \cdot 4=1000 \mathrm{~K}+268 \Rightarrow 116 \mathrm{n}=1000 \mathrm{~K}+268(1)$ | 2 |
| $23 \mathrm{n} \cdot 5=1000 \mathrm{Q}+145 \Rightarrow 115 \mathrm{n}=1000 \mathrm{Q}+145(2)$ | 2 |
| Subtracting relations (1) and (2) we get $\mathrm{n}=1000(\mathrm{~K}-\mathrm{Q})+123$ | 1 |
| The remainder of the division of $\mathrm{n}$ by 1000 is 123. | 1 |