Express the repeating decimal .1405405405405405.... as a

0.1405405405 can be written in fractional form as (1405-1)/9990.

Other examples can be 0.121212121=(121-1)/990

In short, for denominator: place as much nine as repeating number followed by same number of zero as non repeated number.

in numerator: difference between all number formed by all the decimal digit and non repeatative digits.

Hope this help!

Shishir

You know that both Nr and Dr are divisible by 9 (as the sum of the digits in each number is divisble by 9).Now you will get 45/111 now again you know this is divisble by 3.

Anybody knows a good link to a site explaining these division rules.

Alright guys, here is the process:
suppose x=.1405405405405405, which is actually is not a repeating decimal. lets make it as repeting decimal by multiplying 10 (because with this multiplication 1 comes before the decimal and remaining decimal will be the repeating decimal) as under:
(i) 10x=1.405405405405
(ii) multiply this eq by 10^3 (because there are 3 repeating numbers: 4, 0, and 5) => 1,000(10x)=1405.405405405405405405405405405405
(iii) now, substract 10x from both side (because we are eliminating the decimal)

9,990x=1405.405405405-1.405405405

x=1404/9990=26/185

i believe, this method can be applied in any fractions. for example
x=0.2222222222222222222222222222222
(i) x=0.222222222222222
(ii) 10x= 2.2222222222222222
(iii) 9x=2
x =2/9

typical backsolving question
not really concerning

OOOOOOOOOOHHHHHHHHHHHHHHHHHHHHHHHHHHH,

=26/185

I agree with thearch.It is definately backsolvable question.However I would love to know if there is any easy method to solve it.

Great !!! Now that I know the trick, I wish I get a question in the test on this.

Looks like i just saw magic...WOW! But what are the odds of seeing this on the test?

is there any fast way to find the equivalent for : 405/999?

as 15/37?

thank you 700Plus for the tip!!, now I remember reading this on the Kaplan Math workbook....


my bad.......

Yup that's how to treat repeating decimals. Multiply by 10s to get the repeating part and then subtract. That's the logic behind the 9s.

What about other #s? For example 3/11? There are 2 digits repeating, but they are 2 & 7 not 3. Thoughts?

Let x = .1405405...
Then 1000x = 140.5405405...
Subtracting the first equation from the second so that the red portions are eliminated, we get:
\(999x = 140.4\)
\(x = \frac{140.4}{999} = \frac{1404}{9990} = \frac{156}{1110} = \frac{52}{370} = \frac{26}{185}\)

Another example:
Express 0.71212... as a fraction in its most reduced form.

Let x = .71212...
Then 100x = 71.21212...
Subtracting the first equation from the second so that the red portions are eliminated, we get:
\(99x = 70.5\)
\(x = \frac{70.5}{99} = \frac{141}{198} = \frac{47}{66}\)

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