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Convert 0.999897969594......07060504030201999897969594......07060504030201......0.999897969594......07060504030201999897969594......07060504030201...... into the simplest fraction.

Note by Bryan Lee Shi Yang
4 years, 7 months ago

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The given value is simply 0.9998970302010.\overline{999897\ldots 030201}. If it is taken as xx, then we have,

100x=99.98979695020199100x=99.\overline{98979695\ldots 020199}

Performing calculation of 100xx=99x100x-x=99x with finite continuations of the decimals yields the following values:

98.9802 , 98.989803 , 98.98989804 , 98.9898989805 ,  , 98.989898989898 repeated n times (n+1)98.9802~,~98.989803~,~98.98989804~,~98.9898989805~,~\ldots~,~\overline{98.\underbrace{9898989898\ldots}_{98\textrm{ repeated }n\textrm{ times }}(n+1)}

From this, we can conclude that 100xx=99x=98.98100x-x=99x=98.\overline{98}. Now, simplifying this further, we have,

100×99x=9800+99x    99(100xx)=9800    99×99x=9800    x=98009801100\times 99x=9800+99x\implies 99(100x-x)=9800\implies 99\times 99x=9800\implies x=\frac{9800}{9801}

x=98009801\therefore\quad x=\frac{9800}{9801}

Prasun Biswas - 4 years, 7 months ago

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Here's another hint. What is 100xx 100 x - x ?

Calvin Lin Staff - 4 years, 7 months ago

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I wonder whether those hints are for me. :3 Thanks for helping, by the way. I tried simplifying it earlier, but I couldn't find a pattern any helpful pattern.

When you told to look at 1x1-x, I noticed that there's a pattern.

1x=0.0102030496979899=x with reversed decimal component1-x=0.\overline{01020304\ldots 96979899}=x \textrm{ with reversed decimal component}

Since xx is a repeating decimal, I'm not quite sure whether the computation of 1x1-x is correct or not. I was trying to look into it further but then I forgot about it and went to sleep. :D

Well, I'm going to try again. Let's see if I succeed this time. :D

Prasun Biswas - 4 years, 7 months ago

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Typically, the hints are meant for everyone in general. In cases where I reply to a specific person, the hint is meant to guide their way of thinking. Even then, anyone can look at the conversation and work down that path.

You have to be careful when looking at 1x 1 - x . You are no longer working with 1.0000 1.0000 , but actually with 0.99999 0. 99999 . For example, 10.9ˉ0.1ˉ 1 - 0. \bar{9} \neq 0. \bar {1} .

As a further hint to 1x 1 - x , look at Arithmetic-Geometric Progression. It is a simple extension of AP/GP that I like.

Calvin Lin Staff - 4 years, 7 months ago

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@Calvin Lin Sir , there's some guy out here who's mocking you by posting a note and stuff like that . Pls take care of him

A Former Brilliant Member - 4 years, 7 months ago

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Ah, thanks for this one. I found an interesting pattern. The difference is 98.9898.\overline{98} which makes the problem quite easy.

Prasun Biswas - 4 years, 7 months ago

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Dude ,finished studying for Maths ?

A Former Brilliant Member - 4 years, 7 months ago

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@A Former Brilliant Member Well, kind of. I need to revise vector algebra and probability. I'll do that in the next couple of days before 18th. I'm gonna chill for the time being. :D

Prasun Biswas - 4 years, 7 months ago

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@Prasun Biswas Yeah , me too . I have yet to study LPP though !

A Former Brilliant Member - 4 years, 7 months ago

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@A Former Brilliant Member LPP is quite easy. It's simply optimization of functions by graphical methods (atleast for our syllabus). If you're familiar with graphing linear inequations in two variable, LPP in our syllabus is a no brainer.

The only thing to be careful of while doing LPP problems is not to lose focus or be impatient as it may cause one to overlook useful data that was to be made use to form a constraint. And finally, the objective function is computed by corner point theorem on the convex solution set (if there is one).

The only thing that irritates me is the time limit. I'm slow in manual graph plotting.

Prasun Biswas - 4 years, 7 months ago

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@Prasun Biswas Thanks for explaining it to me , I'll look it uo right away to make sure I retain it :)

Thanks again :)

A Former Brilliant Member - 4 years, 7 months ago

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Finally done! Thanks for the help! :D

The final solution is in my original comment. The answer came as 98009801\dfrac{9800}{9801}. I've verified the result with a calculator.

Prasun Biswas - 4 years, 7 months ago

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It would be nice to post this as a problem :)

Joel Tan - 4 years, 7 months ago

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What?

Bryan Lee Shi Yang - 4 years, 7 months ago

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Why do you have periods at the end?Are the numbers supposed to continue into the negative range?

D G - 4 years, 7 months ago

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This is simply a non-terminating recurring decimal. The periods ()(\ldots) indicate that it never terminates.

Prasun Biswas - 4 years, 7 months ago

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Hint: Look at 1x 1 - x .

Extended Hint: Look at Arithmetic-Geometric Progression.

Calvin Lin Staff - 4 years, 7 months ago

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what about 9/10

Vaibhav Prasad - 4 years, 7 months ago

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910=0.90.999897030201\frac{9}{10}=0.9\neq 0.\overline{999897\ldots 030201}

Prasun Biswas - 4 years, 7 months ago

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then how about 999897.......030201100000.......000000\frac { 999897.......030201 }{ 100000.......000000 }

Vaibhav Prasad - 4 years, 7 months ago

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@Vaibhav Prasad That is quite similar to what I proposed in my comment below.

Also, it is a terminating decimal and as such, it cannot be the answer (even if that's the simplest form).

Prasun Biswas - 4 years, 7 months ago

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@Prasun Biswas :3

Vaibhav Prasad - 4 years, 7 months ago

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