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Find integers x,y such that the repeating decimal 0.712341234.... = x/y.

I would actually do this problem if the 7 was not there. If the 7 was not there, my proof would be as follows.

proof:

Let z = 0.12341234...

Then 10^4z = 1234.1234

10^4z - z = 1234

z = 1234/(10^4 - 1)

x = 1234, y = 10^4-1

So my question is, how would this change when there is a random number thrown in there that is not part of the repeating decimal?

Edit: Proof after hints given

Let Let z = 0.712341234...

Then 10z = 7.1234...

10z - 7 = 0.1234...

10^4*(10z - 7) = 1234.1234...

10^4*(10z-7) - (10z-7) = 1234

(10z-7) * 10^4 - 1 = 1234

(10z-7) = 1234/(10^4 - 1)

10z = 1234/(10^4-1) + 7

z = (1234/(10^4-1) + 7)/10

x = 1234/(10^4-1) + 7, y = 10

I mean this does give me the correct answer, but x isn't exactly an integer.

  • 3
    Multiply by $10$, subtract $7$. Later, add $7$, divide by $10$. – 2017-02-09
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    It wouldn't change anything at all. – 2017-02-09
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    z = (1234/(10^4-1) + 7)/10, now simplify the right hand side until you have a fraction, the numerator is x, the denominator is y. – 2017-02-09

3 Answers 3

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We have \begin{align} 0.7\overline{1234} &= \frac{7.\overline{1234}}{10}\\ &= \frac{1}{10}\left(7+0.\overline{1234}\right)\\ &= \frac{1}{10}\left(7+\frac{1234}{9999}\right)\\ &= \frac{1}{10}\left(\frac{7\cdot 9999}{9999} + \frac{1234}{9999}\right)\\ &= \frac{7\cdot 9999 + 1234}{10\cdot 9999}\\ &= \frac{71227}{99990} \end{align} Therefore $x=71227$ and $y=99990$.

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$$0.7\overline{1234} = 0.7 + 0.0\overline{1234} $$

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Let $x=.712341234.... $

$10000x=7123.41234..... $

$9999x =7122.7$

$x =\frac {71227}{99990} $

May be able to reduce. Or maybe not.

The irregular 7 doesn't change anything seriously.

$1/n $ will reduce to a purely repeating pattern but most $m/n $ will not.

...or if you prefer (it's the same thing really)

Let $10x =7.12341234....$

Let $y=.12341234$

$y=\frac {1234}{9999} $

$10x =7+ \frac {1234}{9999} $

$x=\frac 7 {10}+ \frac {1234}{99990} $

$x=\frac {7*9999+1234}{99990}$

$x=\frac {69993+1234}{99990} $

$x=\frac {71227}{99990} $

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    I myself prefer the classic \begin{align}10x&=7.\overline{1234}\\ 100000x&=71234.\overline{1234}\\ 99990x&=71234-7=71227\\ \implies x&=\frac{71227}{99990}\end{align} – 2017-02-09
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    The OP already assignes a special meaning to the variables x an y in his question. I think it is not a good idea to use it with a different meaning in your answer, – 2017-02-09
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    Meh... he asked how to do it. From start. Variables aren't set in stone. (By definition). – 2017-02-09
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    Fimpellizieri, I'm not sure I see the difference. – 2017-02-09