15. [9] Let $a, b, c$ be positive integers such that
$$
\frac{a}{77}+\frac{b}{91}+\frac{c}{143}=1 .
$$

What is the smallest possible value of $a+b+c$ ?
Answer:
79
We need $13 a+11 b+7 c=1001$, which implies $13(a+b+c-77)=2 b+6 c$. Then $2 b+6 c$ must be divisible by both 2 and 13 , so it is minimized at 26 (e.g. with $b=10, c=1$ ). This gives $a+b+c=79$.