## Task 3 - 050623

A natural number $b$ is sought that satisfies the following conditions:

(1) $40<b<600$,

(2) $b$ is divisible by both 4 and 9,

(3) $b$ is not divisible by 8 and not by 27,

(4) $b$ leaves a remainder of 6 when divided by 11.

How many such numbers are there?
Due to (2), $b$ is also divisible by 36. Therefore, 36 must be multiplied by a natural number to obtain $b$.

This factor cannot be 0 or 1 due to (1), and it cannot be divisible by 2 or 3 due to (3). Due to (4), 11 is also ruled out as a factor.

Therefore, due to $600=36 \cdot 16+24$ and the above considerations, only the numbers 5, 7, and 15 remain as possible factors. Thus, only the numbers 180, 252, and 468 need to be examined, all of which satisfy (1), (2), and (3).

Of these, only 468 also satisfies (4). Therefore, $b=468$ is the only solution.