\section*{Exercise 1 - 171211}

Determine all five-digit natural numbers in the decimal positional system that are divisible by 17, 19, and 23, and whose tens digit is 0.
}

Assume a natural number \(a\) has the required properties. Then \(a\) is divisible by the prime numbers 17, 19, and 23, and thus also by their product \(17 \cdot 19 \cdot 23 = 7429\). Therefore, there exists an integer \(n\) such that \(a = 7429 n\).

Since \(a\) is a five-digit number, it follows that \(10000 \leq 7429 n \leq 99999\), hence \(2 \leq n \leq 13\). If \(y\) is the units digit of \(a\), then: Since \(a\) has a tens digit of 0, the number \(a - y\) has zeros in its last two digits, so it is divisible by 1100. Therefore, there exists an integer \(g\) such that \(7429 n - y = 100 g\) or

\[
29 n = 100(g - 74 n) + y
\]

If we set \(x = g - 74 n\), we get \(29 n = 100 x + y\) or \(100 x = 29 n - y\). Given \(2 \leq n \leq 13\) and \(0 \leq y \leq 9\), it follows that

\[
49 = 29 \cdot 2 - 9 \leq 29 n - y = 100 x \leq 29 \cdot 13 - 0 = 377
\]

Thus, \(1 \leq x \leq 3\). From \(29 n = 100 x + y\) or \(29(n - 3 x) = 13 x + y\), if we set \(z = n - 3 x\), we get

\[
13 = 13 \cdot 1 + 0 \leq 29 z = 13 x + y \leq 13 \cdot 3 + 9 = 48
\]

Thus, \(z = 1\).

From this, it follows that \(13 x + y = 29\), so \(29 - y\) is divisible by 13, which, given \(0 \leq y \leq 9\), is only true for \(y = 3\). Therefore, \(x = 2\), \(n = 3 x + z = 7\), and \(a = 7429 \cdot 7 = 52003\). Hence, only this number can have the required properties.

It does have these properties; as a multiple of \(17 \cdot 19 \cdot 23\), it is divisible by 17, 19, and 23, and it has 0 as its tens digit.

Adapted from \([5]\)