Convert 0.999897969594......07060504030201999897969594......07060504030201......0.999897969594......07060504030201999897969594......07060504030201......0.999897969594......07060504030201999897969594......07060504030201...... into the simplest fraction.
Note by Bryan Lee Shi Yang 6 years, 3 months ago
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The given value is simply 0.999897…030201‾0.\overline{999897\ldots 030201}0.999897…030201. If it is taken as xxx, then we have,
100x=99.98979695…020199‾100x=99.\overline{98979695\ldots 020199}100x=99.98979695…020199
Performing calculation of 100x−x=99x100x-x=99x100x−x=99x with finite continuations of the decimals yields the following values:
98.9802 , 98.989803 , 98.98989804 , 98.9898989805 , … , 98.9898989898…⏟98 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)}98.9802 , 98.989803 , 98.98989804 , 98.9898989805 , … , 98.98 repeated n times 9898989898…(n+1)
From this, we can conclude that 100x−x=99x=98.98‾100x-x=99x=98.\overline{98}100x−x=99x=98.98. Now, simplifying this further, we have,
100×99x=9800+99x ⟹ 99(100x−x)=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}100×99x=9800+99x⟹99(100x−x)=9800⟹99×99x=9800⟹x=98019800
∴x=98009801\therefore\quad x=\frac{9800}{9801}∴x=98019800
Here's another hint. What is 100x−x 100 x - x 100x−x?
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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−x1-x1−x, I noticed that there's a pattern.
1−x=0.01020304…96979899‾=x with reversed decimal component1-x=0.\overline{01020304\ldots 96979899}=x \textrm{ with reversed decimal component}1−x=0.01020304…96979899=x with reversed decimal component
Since xxx is a repeating decimal, I'm not quite sure whether the computation of 1−x1-x1−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 1 - x 1−x. You are no longer working with 1.0000 1.0000 1.0000, but actually with 0.99999 0. 99999 0.99999. For example, 1−0.9ˉ≠0.1ˉ 1 - 0. \bar{9} \neq 0. \bar {1} 1−0.9ˉ=0.1ˉ.
As a further hint to 1−x 1 - x 1−x, look at Arithmetic-Geometric Progression. It is a simple extension of AP/GP that I like.
@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
Ah, thanks for this one. I found an interesting pattern. The difference is 98.98‾98.\overline{98}98.98 which makes the problem quite easy.
Dude ,finished studying for Maths ?
@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 – Yeah , me too . I have yet to study LPP though !
@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 – Thanks for explaining it to me , I'll look it uo right away to make sure I retain it :)
Thanks again :)
Finally done! Thanks for the help! :D
The final solution is in my original comment. The answer came as 98009801\dfrac{9800}{9801}98019800. I've verified the result with a calculator.
It would be nice to post this as a problem :)
What?
Why do you have periods at the end?Are the numbers supposed to continue into the negative range?
This is simply a non-terminating recurring decimal. The periods (…)(\ldots)(…) indicate that it never terminates.
Hint: Look at 1−x 1 - x 1−x.
Extended Hint: Look at Arithmetic-Geometric Progression.
what about 9/10
910=0.9≠0.999897…030201‾\frac{9}{10}=0.9\neq 0.\overline{999897\ldots 030201}109=0.9=0.999897…030201
then how about 999897.......030201100000.......000000\frac { 999897.......030201 }{ 100000.......000000 } 100000.......000000999897.......030201
@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 – :3
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
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
The given value is simply 0.999897…030201. If it is taken as x, then we have,
100x=99.98979695…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 , … , 98.98 repeated n times 9898989898…(n+1)
From this, we can conclude that 100x−x=99x=98.98. Now, simplifying this further, we have,
100×99x=9800+99x⟹99(100x−x)=9800⟹99×99x=9800⟹x=98019800
∴x=98019800
Here's another hint. What is 100x−x?
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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=0.01020304…96979899=x 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
Log in to reply
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, 1−0.9ˉ=0.1ˉ.
As a further hint to 1−x, look at Arithmetic-Geometric Progression. It is a simple extension of AP/GP that I like.
Log in to reply
Ah, thanks for this one. I found an interesting pattern. The difference is 98.98 which makes the problem quite easy.
Log in to reply
Dude ,finished studying for Maths ?
Log in to reply
Log in to reply
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.
Log in to reply
Thanks again :)
Finally done! Thanks for the help! :D
The final solution is in my original comment. The answer came as 98019800. I've verified the result with a calculator.
It would be nice to post this as a problem :)
Log in to reply
What?
Why do you have periods at the end?Are the numbers supposed to continue into the negative range?
Log in to reply
This is simply a non-terminating recurring decimal. The periods (…) indicate that it never terminates.
Hint: Look at 1−x.
Extended Hint: Look at Arithmetic-Geometric Progression.
what about 9/10
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109=0.9=0.999897…030201
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then how about 100000.......000000999897.......030201
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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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