## Task 1 - 330631

Find all possible ways to combine three natural numbers $a, b, c$ such that $a+b+c=12$ and $c-b=3$!

Hint:

1. Zero should also be considered a natural number.
2. It is also allowed that among the numbers $a, b, c$, some are equal to each other.
It must be $b \leq 4$; for if $b \geq 5$, then due to $c-b=3$, it would follow that $c \geq 8$.

Thus, $b+c \geq 13$, and therefore $a+b+c \geq 13$, which contradicts $a+b+c=12$.

For $b=0$, it follows that $c=3$ and due to $a+b+c=12$, then $a=9$.

For $b=1$, it follows that $c=4$ and due to $a+b+c=12$, then $a=7$.

For $b=2$, it follows that $c=5$ and due to $a+b+c=12$, then $a=5$.

For $b=3$, it follows that $c=6$ and due to $a+b+c=12$, then $a=3$.

For $b=4$, it follows that $c=7$ and due to $a+b+c=12$, then $a=1$.

Thus, all the desired combinations have been found.