Question

20 points up for the answer

Answer 1

$\dfrac ab=\dfrac38\implies a=\dfrac38b$
$\dfrac bc=\dfrac6{11}\implies c=\dfrac{11}6b$

$\implies a+b+c=\left(\dfrac38+1+\dfrac{11}6\right)b=\dfrac{77}{24}b$

Since each of $a,b,c$ are integers, so must be $a+b+c=\dfrac{77}{24}b$. The only way for this to be the case is if $b$ is a multiple of 24 because 77 and 24 share no common divisors. The smallest multiple would be $b=24$. So the smallest value of $a+b+c$ is 77.