Problem: In base R1 the expanded fraction F1 becomes .373737⋯ and the expanded fraction F2 becomes .737373⋯. In base R2 fraction F1, when expanded, becomes .252525⋯ while fraction F2 becomes .525252⋯. The sum of R1 and R2, each written in the base ten, is:
Answer Choices:
A. 24
B. 22
C. 21
D. 20
E. 19
Solution:
F1=R12−13R1+7=R22−12R2+5 and F2=R12−17R1+3=R22−15R2+2∴2R2+53R1+7=R22−1R12−1=5R2+27R1+3
∴R2=R1+2929R1+1 Knowing that R1 and R2 must each be integral, and that R1⩾8 (why?), we solve for R2 with permissible values of R1. For R1=8,R2 is not integral; for R1=9 or 10,R2 is not integral; for R1=11,R2=8; for R1=12,R2 is not integral; for R1=13,R2=9. The values R1=13,R2=9 do not satisfy the conditions of the problem; the values R1=11,R2=8 do. ∴R1+R2=19