A and B are digits. Since the three digit number 5AB is 40 more than 11 times the two digit number AB, what is the sum of A+B?

Answer 1

We can first simplify the given equation, which is $5AB=40+11(AB)$. We can distribute the 11 to get $5AB=40+11AB$. Subtracting $11AB$ from both sides gives $-6AB=40$. Since we know $A$ and $B$ must be integers, we can determine that $AB$ must be a multiple of $-6$. The only possible values for $AB$ are $-6$ and $-12$, both of which mean that $A$ and $B$ must be negative. Since we are finding the sum $A+B$, the answer is $-1+0=\boxed{-1}$.