1966 AMC 12 Problem 39

Question

In base

R_1

the expanded fraction

F_1

becomes

0.373737\ldots

and the expanded fraction

F_2

becomes

0.737373\ldots.

In base

R_2

fraction

F_1,

when expanded, becomes

0.252525\ldots,

while fraction

F_2

becomes

0.525252\ldots.

The sum of

R_1

and

R_2,

each written in base ten, is:

24

22

21

20

19

A repeating pair

ab

in base

R

equals

\frac{aR+b}{R^2-1}

Add the equations for

F_1

and

F_2,

then subtract them

The two descriptions give

\begin{aligned} F_1&=\frac{3R_1+7}{R_1^2-1}\\ &=\frac{2R_2+5}{R_2^2-1},\\ F_2&=\frac{7R_1+3}{R_1^2-1}\\ &=\frac{5R_2+2}{R_2^2-1}. \end{aligned}

Adding yields

\frac{10}{R_1-1}=\frac{7}{R_2-1},

or

10R_2-7R_1=3.

Subtracting yields

\frac{4}{R_1+1}=\frac{3}{R_2+1},

or

4R_2-3R_1=-1.

Solving gives

R_1=11,

R_2=8,

whose sum is

19.

Therefore, the correct answer is E.

Problem 39 in other years