Question

20 points up for the answer

Answer 1

`\frac ab=\frac38\implies a=\frac38b`

`\frac bc=\frac6{11}\implies c=\frac{11}6b`


`\implies a+b+c=\left(\frac38+1+\frac{11}6\right)b=(77)/(24)b`

Since each of
`a,b,c` are integers, so must be
`a+b+c=(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.