Problem 3. Consider the prime numbers $p<q<r$ and the non-zero natural number $a<p$. The number $n$ is the smallest non-zero natural number which, when divided by $p, q$, and $r$, gives a remainder that is $a$ less than the divisor each time.

a) Determine, in terms of $p, q, r$, and $a$, the number $n$.

b) If $n=1000$, determine the possible values of $a$.
Solution: a) From the statement $n=p c_{1}+p-a$ or $n+a=p\left(c_{1}+1\right)$, from which we deduce that $p \mid n+a$.

Similarly, we deduce $q \mid n+a$ and $r \mid n+a \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

Since $p, q, r$ are different prime numbers, it follows that $p q r \mid n+a$, from which

![](https://cdn.mathpix.com/cropped/2024_06_07_37e0bde509fee30fee69g-4.jpg?height=52&width=1260&top_left_y=874&top_left_x=427)
b) $n=1000$ implies $p q r-a=1000$. From $a<p$ it follows that $p q r-p<1000$ or $p(q r-1)<1000$. ................................................................................

If $p \geq 11$, then $q \geq 13$ and $r \geq 17$, from which $p(q r-1) \geq 11 \cdot(13 \cdot 17-1)$, that is $p(q r-1) \geq 1000$. We deduce that $p \in\{2,3,5,7\} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ p

Since $a<p$ we have $a \in\{1,2,3,4,5,6\}$. If $a=1$ we obtain the solution

![](https://cdn.mathpix.com/cropped/2024_06_07_37e0bde509fee30fee69g-4.jpg?height=60&width=1260&top_left_y=1152&top_left_x=427)

Justification that the other cases are not possible. ............. 1 p