[Algebra]
Problem: The rational number $r$ is the largest number less than 1 whose base-7 expansion consists of two distinct repeating digits, $r=0 . \underline{A} \underline{B} \underline{A} \underline{B} \underline{A} \underline{B} \ldots$ Written as a reduced fraction, $r=\frac{p}{q}$. Compute $p+q$ (in base 10).

Solution: In base 7, the value of $r$ must be $0.656565 \ldots=0 . \overline{65}_{7}$. Then $100_{7} \cdot r=65 . \overline{65}_{7}$, and $\left(100_{7}-1\right) r=$ $65_{7}$. In base $10,65_{7}=6 \cdot 7+5=47_{10}$ and $100_{7}-1=7^{2}-1=48_{10}$. Thus $r=47 / 48$, and $p+q=95$.

Final answer: 95