If $a$ and $b$ are positive integers and $\frac{20}{19}=1+\frac{1}{1+\frac{a}{b}}$, what is the least possible value of $a+b$ ?
(A) 16
(B) 19
(C) 20
(D) 38
(E) 39

Part C: Each correct answer is worth 8.
We begin by expressing $\frac{20}{19}$ in a form that is similar to the right side of the given equation. Converting $\frac{20}{19}$ to a mixed fraction we get, $\frac{20}{19}=1 \frac{1}{19}=1+\frac{1}{19}$.

Since $\frac{20}{19}=1+\frac{1}{1+\frac{a}{b}}$ and $\frac{20}{19}=1+\frac{1}{19}$, then $1+\frac{1}{1+\frac{a}{b}}=1+\frac{1}{19}$ and so $\frac{1}{1+\frac{a}{b}}=\frac{1}{19}$.

The numerators of $\frac{1}{1+\frac{a}{b}}$ and $\frac{1}{19}$ are each equal to 1 , and since these fractions are equal to one another, their denominators must also be equal.

That is, $1+\frac{a}{b}=19$ and so $\frac{a}{b}=18$.

Since $a$ and $b$ are positive integers, then the fractions $\frac{a}{b}$ which are equal to 18 are $\frac{18}{1}, \frac{36}{2}, \frac{54}{3}$, and so on.

Thus, the least possible value of $a+b$ is $18+1=19$.

ANSWER: (B)