\section*{Exercise 1 - 091031}

Give all three-digit natural numbers divisible by 11 that, when divided by 5, leave a remainder of 1 and, when divided by 7, leave a remainder of 3!
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Let \(n\) be such a number. Then, with \(n\), \(n-66=n-11 \cdot 6\) is also divisible by 11. If \(n\) leaves a remainder of 1 when divided by 5, then \(n-66=(n-1)-5 \cdot 13\) is also divisible by 5. And finally: If \(n\) leaves a remainder of 3 when divided by 7, then \(n-66=(n-3)-7 \cdot 9\) is also divisible by 7.

Since 5, 7, and 11 are pairwise coprime, \(n-66\) is therefore divisible by the product \(5 \cdot 7 \cdot 11=385\), so one can represent \(n\) as \(n=385 \cdot k+66\) with a non-negative integer \(k\).

Clearly, for \(k \geq 3\), \(n>1000\) as well, and thus no longer a three-digit number (and for \(k=0\) only a two-digit number), so one obtains exactly the following two solutions:

\(n_{1}=385 \cdot 1+66=451\) and \(n_{2}=385 \cdot 2+66=836\).