Here's another hint. What is \( 100 x - x \)?
–
Calvin Lin
Staff
·
2 years ago

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@Calvin Lin
–
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.
–
Prasun Biswas
·
2 years ago

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@Calvin Lin
–
Ah, thanks for this one. I found an interesting pattern. The difference is \(98.\overline{98}\) which makes the problem quite easy.
–
Prasun Biswas
·
2 years ago

@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
–
Prasun Biswas
·
2 years ago

@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.
–
Prasun Biswas
·
2 years 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 :)

@Calvin Lin
–
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
–
Prasun Biswas
·
2 years ago

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@Prasun Biswas
–
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, \( 1 - 0. \bar{9} \neq 0. \bar {1} \).

@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
–
Azhaghu Roopesh M
·
2 years ago

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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,

## Comments

Sort by:

TopNewestIt would be nice to post this as a problem :) – Joel Tan · 2 years ago

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– Bryan Lee Shi Yang · 2 years ago

What?Log in to reply

Here's another hint. What is \( 100 x - x \)? – Calvin Lin Staff · 2 years ago

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The final solution is in my original comment. The answer came as \(\dfrac{9800}{9801}\). I've verified the result with a calculator. – Prasun Biswas · 2 years ago

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– Prasun Biswas · 2 years ago

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

– Azhaghu Roopesh M · 2 years ago

Dude ,finished studying for Maths ?Log in to reply

– Prasun Biswas · 2 years ago

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. :DLog in to reply

– Azhaghu Roopesh M · 2 years ago

Yeah , me too . I have yet to study LPP though !Log in to reply

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 · 2 years ago

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Thanks again :) – Azhaghu Roopesh M · 2 years ago

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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 – Prasun Biswas · 2 years ago

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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. – Calvin Lin Staff · 2 years ago

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– Azhaghu Roopesh M · 2 years ago

Sir , there's some guy out here who's mocking you by posting a note and stuff like that . Pls take care of himLog in to reply

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,

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

\[\therefore\quad x=\frac{9800}{9801}\] – Prasun Biswas · 2 years ago

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what about 9/10 – Vaibhav Prasad · 2 years ago

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– Prasun Biswas · 2 years ago

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

– Vaibhav Prasad · 2 years ago

then how about \(\frac { 999897.......030201 }{ 100000.......000000 } \)Log in to reply

Also, it is a terminating decimal and as such, it cannot be the answer (even if that's the simplest form). – Prasun Biswas · 2 years ago

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– Vaibhav Prasad · 2 years ago

:3Log in to reply

Hint:Look at \( 1 - x \).Extended Hint:Look at Arithmetic-Geometric Progression. – Calvin Lin Staff · 2 years agoLog in to reply

Why do you have periods at the end?Are the numbers supposed to continue into the negative range? – D G · 2 years ago

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– Prasun Biswas · 2 years ago

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