Repeating decimal, part 2

Number Theory Level 2

$0.\overline{00010203040506\ldots 969799}=\frac{1}{N}$

Given that $N$ is a 4-digit integer, find $N$ .

Clarification : The repeated digits are from 00, 01, 02, ..., till 99, without 98 , the period is 198.

The answer is 9801.

4 solutions

Darryl Stein
Dec 7, 2017

$\frac{1}{N}$ = $\frac{1}{10000}$ + $\frac{2}{1000000}$ + $\frac{3}{100000000}$ + ... + $\frac{97}{10^{196}}$ + $\frac{99}{10^{198}}$

$\frac{100}{N}$ = $\frac{1}{100}$ + $\frac{2}{10000}$ + $\frac{3}{1000000}$ + ... + $\frac{97}{10^{194}}$ + $\frac{99}{10^{196}}$

$\frac{99}{N}$ = $\frac{1}{100}$ + $\frac{1}{10000}$ + $\frac{1}{1000000}$ + ... + $\frac{1}{10^{194}}$ + $\frac{2}{10^{196}}$ - $\frac{99}{10^{198}}$

$\frac{99}{N}$ = $\dfrac{\frac{1}{100}} {1 - \frac{1}{100}}$

$\frac{99}{N}$ = $\frac{1}{99}$

$\frac{1}{N}$ = $\frac{1}{9801}$

Did you just forgot the repeated digits? Are you also assuming that the subtracted term doesn't make much difference? (Just asking)

Akshay Krishna - 2 years, 6 months ago

Scott Rodham
Mar 7, 2017

As an estimate 10^10÷1020304 = 9801.000486

X X
Jul 21, 2018

This equals $(\dfrac1{100}+\dfrac1{100^2}+\dfrac1{100^3}+\cdots)^2=(\dfrac1{99})^2=\dfrac1{9801}$

Can you explain why does this work ignoring the last few digits before the bar?

Akshay Krishna - 2 years, 6 months ago

Bert Seegmiller
Mar 14, 2018

Divided 1 by the 1st 29 digits of the fraction.

Repeating decimal, part 2

The answer is 9801.

4 solutions

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