8. For three different positive integers $a, b$, and $c$, it holds that $a+2b+3c<12$. Which of the following inequalities is certainly satisfied?
A) $3a+2b+c<17$
B) $a+b+c<7$
C) $a-b+c<4$
D) $b+c-a<3$
E) $3b+3c-a<6$

## B-questions
A8. D) $b+c-a<3$ We first determine all possible solutions $(a, b, c)$ of the inequality $a+2 b+3 c<12$, with $a$, $b$ and $c$ being different positive integers. From $3 c<12-a-2 b \leqslant 9$ it follows that $c<3$, so $c=2$ or $c=1$.

- Case $c=2$ We have $a+2 b<6$, so $b<3$. Since $b$ must not be equal to $c$, $b$ must be 1. We now find exactly one solution: $(a, b, c)=(3,1,2)$.
- Case $c=1$ We have $2 b<9-a \leqslant 8$, so $b<4$. If $b=3$, then $a<3$ and not equal to 1, so we only find the solution $(a, b, c)=(2,3,1)$. If $b=2$, then $a<5$, so $a=3$ or $a=4$, since $a$ must not be equal to 1 or 2. We now find two solutions $(3,2,1)$ and $(4,2,1)$.

The only solutions $(a, b, c)$ are thus $(3,1,2),(2,3,1),(3,2,1)$ and $(4,2,1)$. Since $(4,2,1)$ does not satisfy the inequalities in A) and B), those answers are eliminated. Since $(3,1,2)$ does not satisfy the inequalities in C) and E), those answers are also eliminated. However, all four solutions do satisfy the inequality in D), so that answer is the correct one.
The seven smallest numbers by which 707070 is divisible are $1,2,3,5,6,7,10$. The seventh number in the list is thus $\frac{707070}{10}=70707$.