Fun with Fractions!

Suppose the decimal expansion of a certain fraction a b \frac{a}{b} takes the form

0. 0102030405...979900 0.\overline{0102030405... 979900}

wherein the series of digits skip 98, and then recurrs. If a a and b b are coprime positive integers, determine a + b a+b .


The answer is 9901.

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2 solutions

Kelvin Hong
Nov 24, 2017

Let the answer be S S

S = 1 1 0 2 + 2 1 0 4 + 3 1 0 6 + 4 1 0 8 + . . . S = \frac{1}{10^2} + \frac{2}{10^4} + \frac{3}{10^6} + \frac{4}{10^8} + ...

1 100 S = 1 1 0 4 + 2 1 0 6 + 3 1 0 8 + . . . \frac{1}{100}S = \frac{1}{10^4} + \frac{2}{10^6} + \frac{3}{10^8} + ...

Subtracting, we get

99 100 S = 1 99 \frac{99}{100}S = \frac{1}{99}

S = 100 9801 S = \frac{100}{9801}

There is a little wrong , the center part should be . . . 979900010202... ...979900010202...

Yes, the reports in the problem specifically pointed out the omission of the term 98 in the decimal expansion.

Juvin Abayon - 3 years, 6 months ago
Daniel Xiang
Feb 27, 2018

Let f ( x ) = 1 + 2 x + 3 x 2 + f(x) = 1+2x+3x^2+\cdots , then 1 0 2 f ( 1 0 2 ) = 0. 0102030405...979900 10^{-2}f\left(10^{-2}\right) = 0.\overline{0102030405... 979900}

98 98 is skiped since the 10 0 t h 100^{\mathrm{th}} term carries 1 1

0 x f ( t ) d t = x + x 2 + x 3 + = x 1 x \displaystyle \int^{x}_0 f(t)\mathrm dt = x+x^2+x^3+\cdots = \frac{x}{1-x}

f ( x ) = d d x x 1 x = 1 ( 1 x ) 2 \displaystyle f(x)=\frac{\mathrm d}{\mathrm dx}\frac{x}{1-x} = \frac{1}{(1-x)^2}

1 0 2 f ( 1 0 2 ) = 100 9801 \displaystyle 10^{-2}f\left(10^{-2}\right) = \boxed{ \displaystyle \frac{100}{9801}}

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