Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$, and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$, where $m$, $n$, and $p$ are positive integers and $m$ is not divisible by the square of any prime. Find $100m+10n+p$.
Let the second term of each series be $x$. Then, the common ratio is $\frac{1}{8x}$, and the first term is $8x^2$. 
So, the sum is $\frac{8x^2}{1-\frac{1}{8x}}=1$. Thus, $64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}$. 
The only solution in the appropriate form is $x = \frac{\sqrt{5}-1}{8}$. Therefore, $100m+10n+p = \boxed{518}$.