4. Determine all natural numbers of the form $\overline{9 a 6 b 9}$ divisible by 3, where the digits in the tens and thousands places are prime numbers.
4. A number is divisible by 3 if the sum of its digits is divisible by 3, so the sum $9+a+6+b+9=24+a+b$ is divisible by 3.

1 POINT

Since $a$ and $b$ are prime numbers, then $a, b \in \{2,3,5,7\}$.

For $a=2$, we have that $26+b$ is divisible by 3, so $b=7$.

1 POINT

For $a=3$, we have that $27+b$ is divisible by 3, so $b=3$.

1 POINT

For $a=5$, we have that $29+b$ is divisible by 3, so $b=7$.

1 POINT

For $a=7$, we have that $31+b$ is divisible by 3, so $b \in \{2,5\}$.

1 POINT

The desired numbers are 92679, 93639, 95679, 97629, 97659.

1 POINT

TOTAL 6 POINTS

Note: Correct answers without justification earn 4 points.