# 1966 AHSME Problems/Problem 39

## Problem

In base $R_1$ the expanded fraction $F_1$ becomes $.373737\cdots$, and the expanded fraction $F_2$ becomes $.737373\cdots$. In base $R_2$ fraction $F_1$, when expanded, becomes $.252525\cdots$, while the fraction $F_2$ becomes $.525252\cdots$. The sum of $R_1$ and $R_2$, each written in the base ten, is:

$\text{(A) } 24 \quad \text{(B) } 22 \quad \text{(C) } 21 \quad \text{(D) } 20 \quad \text{(E) } 19$

## Solution

First, let's write $F_1$ as a proper fraction in base $R_1$. To do that, note that: $F_1=0.373737\dots$ Multiplying this equation on both sides $R_1^2$, we get: $R_1^2F_1=37.373737\dots$ Subtracting the first equation from the second one, we get: $R_1^2F_1-F_1=37\\F_1(R_1^2-1)=37\\F_1=\frac{3R_1+7}{R_1^2-1}$ Using a very similar method as above, we can see that: $F_1=\frac{3R_1+7}{R_1^2-1}=\frac{2R_2+5}{R_2^2-1}$ and $F_2=\frac{7R_1+3}{R_1^2-1}=\frac{5R_2+2}{R_2^2-1}$. Dividing the 2 equations to get some potential solutions, we get (note that we don't have to worry about division by zero because $R_1,R_2>7$): $\frac{3F_1+7}{7F_1+3}=\frac{2F_2+5}{5F_2+2}\\15F_1F_2+6F_1+35F_2+14=14F_1F_2+35F_1+6F_2+15\\F_1F_2-29F_1+29F_2=1\\\left(F_1+29\right)\left(F_2-29\right)=-840$ Since non-integer bases are rarely used, we can try to assume that the solution is positive integers and see if we get any solutions. Notice that $(F_2-29)$ has to be a negative factor of 840. We need to plug in values of $F_2 > 7$. 840 divides -21, so we plug in 8 to check. Luckily, when $F_2 = 8$, we see that $F_1=11$, and furthermore, 11+8=19 is one of the answers. We can quickly test it into the original equations $F_1=\frac{3R_1+7}{R_1^2-1}=\frac{2R_2+5}{R_2^2-1}$ and $F_2=\frac{7R_1+3}{R_1^2-1}=\frac{5R_2+2}{R_2^2-1}$, and we see they indeed work. Therefore the answer is $\boxed{E}$