@Azhaghu Roopesh M
–
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

@Azhaghu Roopesh M
–
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.

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 \(1-x\), I noticed that there's a pattern.

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

Since \(x\) is a repeating decimal, I'm not quite sure whether the computation of \(1-x\) 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

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 \( 1 - x \). You are no longer working with \( 1.0000 \), but actually with \( 0. 99999 \). For example, \( 1 - 0. \bar{9} \neq 0. \bar {1} \).

The given value is simply \(0.\overline{999897\ldots 030201}\). If it is taken as \(x\), then we have,

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

Performing calculation of \(100x-x=99x\) with finite continuations of the decimals yields the following values:

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

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

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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

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What?

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Here's another hint. What is \( 100 x - x \)?

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Finally done! Thanks for the help! :D

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

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

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Dude ,finished studying for Maths ?

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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.

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Thanks again :)

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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 \(1-x\), I noticed that there's a pattern.

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

Since \(x\) is a repeating decimal, I'm not quite sure whether the computation of \(1-x\) 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

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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 \( 1 - x \). You are no longer working with \( 1.0000 \), but actually with \( 0. 99999 \). For example, \( 1 - 0. \bar{9} \neq 0. \bar {1} \).

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

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The given value is simply \(0.\overline{999897\ldots 030201}\). If it is taken as \(x\), then we have,

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

Performing calculation of \(100x-x=99x\) with finite continuations of the decimals yields the following values:

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

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

\[100\times 99x=9800+99x\implies 99(100x-x)=9800\implies 99\times 99x=9800\implies x=\frac{9800}{9801}\]

\[\therefore\quad x=\frac{9800}{9801}\]

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what about 9/10

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\[\frac{9}{10}=0.9\neq 0.\overline{999897\ldots 030201}\]

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

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

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Hint:Look at \( 1 - x \).Extended Hint:Look at Arithmetic-Geometric Progression.Log in to reply

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

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

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