## Task 4 - 090524

Determine all natural numbers $z$ for which the following conditions hold simultaneously:

(a) $z$ is odd;

(b) $z$ is divisible by 3, 5, and 7;

(c) $500<z<1000$.
The numbers 3, 5, and 7 are pairwise coprime. Therefore, because \(3 \cdot 5 \cdot 7 = 105\), a number is divisible by 3, 5, and 7 if and only if it is an integer multiple of 105.

Now we have

\[
\begin{gathered}
4 \cdot 105 = 420 \quad, \quad 5 \cdot 105 = 525 \quad, \quad 6 \cdot 105 = 630 \quad, \quad 7 \cdot 105 = 735 \\
8 \cdot 105 = 840 \quad, \quad 9 \cdot 105 = 945 \quad, \quad 10 \cdot 105 = 1050
\end{gathered}
\]

The conditions (b) and (c) are therefore satisfied by the numbers 525, 630, 735, 840, 945, and only by these; because every multiple of 105 with a smaller integer than 5 is less than 500 and every multiple of 105 with a larger integer than 9 is greater than 1000.

Of these numbers, 525, 735, and 945, and only these, also satisfy condition (3). The sought numbers are therefore 525, 735, and 945, and only these.

Solutions of the II. Round 1969 taken from [5]

### 2.11 X. Olympiad 1970

### 2.11.1 I. Round 1970, Class 5