10.9090909091.....

I don't really know what to do with this problem, I know that I should set it up like this:

But whats next?

So A=1090/100 and R=1/100?

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- October 13th 2008, 10:28 AMredman223Series with repeating decimal
10.9090909091.....

I don't really know what to do with this problem, I know that I should set it up like this:

But whats next?

So A=1090/100 and R=1/100? - October 13th 2008, 11:54 AMmr fantastic
- October 13th 2008, 12:31 PMredman223
Will that always work? Do you basically just take the original repeating number s, multiply it by 100, subtract s from it to make it 99s and then divide by 99?

Also, the number was 10.9090909091, the 1 at the end is throwing me off, because I don't know if that trick still applies or not. - October 13th 2008, 12:41 PM11rdc11
The trick is to count how many repeating decimals you have,

For instance in

Muliply both sides of this equation by , where n is the number of digits in the repeating string. Here, the single digit 4 repeats infinitely, so n = 1

Subtract the original equation (x=0.9444444...) from the modified equation (10x = 9.444444...) which equals

9x= 8.5

- October 13th 2008, 01:45 PMredman223
Ok, that makes sense. But what about the number in my problem, I am confused by the way its written.

10.9090909091... I don't know what is repeated. Is it the 1? Is is the whole thing? The problem is that there is no teacher or anything to help me clarify that problem.

So I will have to do it multiple times until the computer accepts the answer.

So would I take 10.9090909091 and multiply it by what? By 1? Or by 10? - October 13th 2008, 02:04 PMPlato
But you may be required to use a series solution.

- October 13th 2008, 02:16 PMredman223
I am still stuck on what to do, because of the 1 at the end of

10.9090909091 - October 13th 2008, 02:53 PMPlato
- October 13th 2008, 02:55 PMWantMath
I think so, too, or they rounded the number by accident

- October 13th 2008, 03:12 PMredman223
Express http://webwork.csufresno.edu/webwork...04dfea90d1.png as a rational number, in the form http://webwork.csufresno.edu/webwork...fce30f9be1.png

where http://webwork.csufresno.edu/webwork...da050b4dd1.png and http://webwork.csufresno.edu/webwork...12a8ea5921.png are positive integers with no common factors.

This is how the problem is written out on webwork (I despise webwork, I wish our school would stop using it). I tried using 1080 and 99 but those didn't work. - October 13th 2008, 04:13 PM11rdc11
- October 13th 2008, 04:29 PMredman223
I tried that earlier and it didn't work. I am stumped.

- October 13th 2008, 04:32 PM11rdc11
- October 13th 2008, 04:34 PMmr fantastic
- October 13th 2008, 04:49 PMmr fantastic
If it's not a typo in the question here is an outline of the ridiculous solution:

10 + 0.909090909 + 0.0000000001111........ (I assume)

Obviously 0.909090909 = 909090909/1000000000

Let S = 0.00000000011111.....

Then 1000000000S = 0.1111111 and 10000000000S = 1.1111111

Therefore 9000000000S = 1 => S = 1/9000000000.

So your number is 10 + (909090909/1000000000) + (1/9000000000).

Now add the fractions in the usual way to get the required representation.