Problem: Let F=.48181⋯ be an infinite repeating decimal with the digits 8 and 1 repeating. When F is written as a fraction in lowest terms, the denominator exceeds the numerator by
Answer Choices:
A. 13
B. 14
C. 29
D. 57
E. 126
Solution:
F=.4+.081+.00081+⋯=.4+.081(1+.01+.0001+⋯)
=.4+.081×1−.011​=(4/10)+(81/1,000)(1/.99)=53/110.
Denominator - Numerator =110−53=57.