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

Question

Express the repeating decimal .1405405405405405.... as a fraction in lowest terms.

anirban16 · Accepted Answer

Well the trick here is that any repeating decimal can be represented as a fraction by taking the Nr as the repeated part and the Dr as the number of 9s as the number of digits that gets repeated.

So say for instance .4545454545.... = 45/99 

.405405405.......... = 405/999

So in our case we have .1405405405...

So multiply by 10 and divide by 10.

So we have 1.405405405.../10 = (1 + .405405405405)/10

= (1 + 405/999)/10 = (1+15/37)/10 = 52/(10*37) = 26/185

anirban16 · Answer

Well it is part of arithmetic. GMAT doesn't have a rigid syllabus for quant, does it?

thearch · Answer

typical backsolving question
not really concerning

MA · Answer

OOOOOOOOOOHHHHHHHHHHHHHHHHHHHHHHHHHHH,

=26/185

700Plus · Answer

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

700Plus · Answer

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

Folaa3 · Answer

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

shishirgupta77 · Answer

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

kevinw · Answer

is there any fast way to find the equivalent for : 405/999?

as 15/37?

700Plus · Answer

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.

MA · Answer

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

kevinw · Answer

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

anirban16 · Answer

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.

jimmyjamesdonkey · Answer

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

GMATGuruNY · Answer

Let x = .1 405405... 
Then 1000x = 140.5 405405... 
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 = .7 1212... 
Then 100x = 71.2 1212... 
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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Express the repeating decimal .1405405405405405.... as a

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