## Task 3 - 200623

Determine all natural numbers $\mathrm{z}$ for which $1000 \leq z \leq 1700$ holds and which are divisible by 9, 12, and 14!
A number is divisible by 9, 12, and 14 if and only if it is divisible by the least common multiple (LCM) of these numbers. Given the prime factorizations

$$
9=3^{2}, \quad 12=2^{2} \cdot 3, \quad 14=2 \cdot 7
$$

this LCM is the number $2^2 \cdot 3^2 \cdot 7 = 252$.

Thus, a natural number $z$ is divisible by 9, 12, and 14 if and only if there is a natural number $n$ such that $z = 252 \cdot n$. All the desired numbers $z$ are therefore obtained from those $n$ for which $1000 \leq 252 n \leq 1700$ holds. Now we observe:

For $n=3$, $252 n = 252 \cdot 3 = 756 < 1000$;

for this $n$, no numbers $z$ are obtained that satisfy the inequality conditions.

For $n=4$, $z = 252 \cdot 4 = 1008$;

for $n=5$, $z = 252 \cdot 5 = 1260$;

for $n=6$, $z = 252 \cdot 6 = 1512$;

these three numbers satisfy the condition $1000 \leq z \leq 1700$.

Therefore, the numbers 1008, 1260, and 1512 are the desired numbers.