Help needed with whole numbers

I need help on this. I tried simplifying it, putting some numbers on it but there has to be a simple way to figure this one out.

a, b and c are whole numbers.

$$a > b > c$$ $$(a + 2b)/c = b$$ what is the lowest number that a can get?

I simplified into something like this but can't go any further

$$c - (a/b) = 2$$

And here are the choices: 8, 9, 10, 11, 12

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You can rewrite the equality $\frac{a+2b}{c}=b$ to the equality $a=b(c-2)$. Then you see that $c$ must be at least $4$, otherwise $b(c-2)$ is not bigger than $b$. It follows that $b$ is at least $5$ and hence $a$ is at least $10$.

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Hint: rewrite $(a+2b)/c = b$ as $a = (c-2)b$. Then consider all possibilities of factoring $a = 8,9,10,11,12$.

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A more useful simplification is $a=b(c-2)$, which you get by multiplying both sides by $c$ and subtracting 2b. so if $a=8$ then the factorization must be 4x2, since $b<8$ and $c-2<7-2$, thus 8x1 is impossible. If $b=4$ then $c=4$, or if $b=2$ then $c=6$, neither of which are allowed. So it can't be 8.

Can you do something similar for the other numbers?

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