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# 0.99999999/1.0001 - 0.99999991/1.0003 =

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26 Dec 2012, 06:36
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$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) $$10^{(-8)}$$

(B) $$3*10^{(-8)}$$

(C) $$3*10^{(-4)}$$

(D) $$2*10^{(-4)}$$

(E) $$10^{(-4)}$$

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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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26 Dec 2012, 06:39
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$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}$$

Now apply $$a^2-b^2=(a+b)(a-b)$$:

$$\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}=\frac{(1+10^{-4})(1-10^{-4})}{1+10^{-4}}-\frac{(1+3*10^{-4})(1-3*10^{-4})}{1+3*10^{-4}}=(1-10^{-4})-(1-3*10^{-4})=2*10^{-4}$$.

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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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22 Jul 2013, 21:16
166
59
$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

Responding to a pm:
To be honest, I can't think of an alternative method. The fractions are really complicated and need to be simplified before proceeding. For simplification, I think you will need to use a^2 - b^2 = (a - b)(a + b)

All I can suggest is that you can try to solve it without the exponents if that seems easier e.g.

$$\frac{0.99999999}{1.0001}-\frac{.99999991}{1.0003}$$

$$\frac{{1 - .00000001}}{{1 + .0001}}-\frac{{1 - .00000009}}{{1 + .0003}}$$

$$\frac{{1^2 - .0001^2}}{{1 + .0001}}-\frac{{1^2 - 0.0003^2}}{{1 + .0003}}$$

$$\frac{{(1 - .0001)(1 + .0001)}}{{(1 + .0001)}}-\frac{{(1 - .0003)(1 + .0003)}}{{(1 + .0003)}}$$

$$(1 - .0001) - (1 - .0003)$$

$$.0002 = 2*10^{-4}$$
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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04 Mar 2016, 11:15
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$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

2 minutes ain't gonna do this math. 3mins may with a high margin of error under exam condition.

Look at the options closely

A. 0.00000001
B. 0.00000003
C. 0.0003
D. 0.0002
E. 0.0001

Only one option is an even number, the rest odd (an esoteric sort of even/odd number)

both fractions in the question are odds
odd minus odd is always even.
Only D is even.

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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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02 Jul 2013, 00:16
4
5
$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

A. 10^-8
B. 3(10^-8)
C. 3(10^-4)
D. 2(10^-4)
E. 10^-4[/quote]

Well, you need a little bit of algebra here...

The question is meant to test the formula $$a^2-b^2=(a+b)(a-b).$$
The clue is in the answers, explicitly the powers of $$10$$.
You can write the given expression as follows:

$$\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}=\frac{(1+10^{-4})(1-10^{-4})}{1+10^{-4}}-\frac{(1+3*10^{-4})(1-3*10^{-4})}{1+3*10^{-4}}=1-10^{-4}-(1-3*10^{-4})=2*10^{-4}.$$

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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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22 Jul 2013, 22:00
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$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

The best solution is already outlined by Bunuel/Karishma. Because this is an Official Problem, I was sure there might be another way to do this.
So i did spend some time and realized that 0.99999999 might be a multiple of 1.0001 because of the non-messy options and found this :

9*1.0001 = 9.0009 ; 99*1.0001 = 99.0099 and as because the problem had 9 eight times, we have 9999*1.0001 = 9999.9999.
Again, looking for a similar pattern, the last digit of 0.99999991 gave a hint that maybe we have to multiply by something ending in 7, as because we have 1.0003 in the denominator. And indeed 9997*1.0003 = 9999.9991. Thus, the problem boiled down to $$9999*10^{-4} - 9997*10^{-4} = 2*10^{-4}$$

D.

Maybe a bit of luck was handy.
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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22 Jul 2013, 23:25
32
7
Here is my alternative solution for this problem (not for all problems):

$$\frac{A}{B} - \frac{C}{D}= \frac{(AD-BC)}{BD}$$.

So $$\frac{0.99999999}{1.0001} - \frac{0.99999991}{1.0003}= \frac{(0.99999999*1.0003-0.99999991*1.0001)}{(1.0001*1.0003)}$$.

For this case, the ultimate digit of 0.99999999*1.0003-0.99999991*1.0001 is 6
In the denominator, the ultimate digit of 1.0001*1.0003 is 3
Therefore, the ultimate digit of the final result is 2. So it should be 2 * 0.00...01 --> Only D has the last digit of 2.

Alternatively, we can calculate each fraction, $$\frac{0.99999999}{1.0001}$$ has last digit of 9, and $$\frac{0.99999991}{1.0003}$$ has last digit of 7, so the final last digit is 2 --> D

This is a special problem. For example $$\frac{...6}{4}$$ can have a result of ...4 or ...9. Therefore, in this case we have to calculate as Bunuel did.

In general, we can only apply this strategy only if the last digit of divisor is 1, 2 or 3.
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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09 Oct 2013, 16:50
How did we even know to apply a^2 - b^2 = (a - b)(a + b) to this problem? I understand the math, but if I saw this problem on the test I would have never guessed to apply that method.

Thanks,
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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09 Oct 2013, 20:36
1
runningguy wrote:
How did we even know to apply a^2 - b^2 = (a - b)(a + b) to this problem? I understand the math, but if I saw this problem on the test I would have never guessed to apply that method.

Thanks,
C

(a^2 - b^2) is the "mathematical" method i.e. a very clean solution that a Math Prof will give you. With enough experience a^2 - b^2 method will come to you. But since most of us are not Math professors, we could get through using brute force. Two alternative approaches have been given by mau5 and lequanftu26. You may want to give them a thorough read.
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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01 Mar 2014, 22:48
10
I just rounded it up, did some questionable math and got lucky, it would seem.

1/1.0001 - 1/1.0003 = ?
1/1.0001 = 1/1.0003
1.0003(1) = 1.0001(1)
1.0003-1.0001=?
0.0002

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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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14 May 2015, 06:01
6
5
I'm going to repost Ron's solution

OR
Just do what a fifth grader would do!

Attachment:

00000370.png [ 81.02 KiB | Viewed 36247 times ]

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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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10 Apr 2016, 09:51
8
4
$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

Another approach is to combine the fractions and then use some approximation.

First combine the fractions by finding a common denominator.
(9999.9999)/(10001) - (9999.9991)/(10003)
= (9999.9999)(10003)/(10001)(10003) - (9999.9991)(10001) /(10003)(10001)
= [(10003)(9999.9999) - (10001)(9999.9991)] / (10001)(10003)
= [(10003)(10^4) - (10001)(10^4)] / (10^4)(10^4) ... (approximately)
= [(10003) - (10001)] / (10^4) ... (divided top and bottom by 10^4)
= 2/(10^4)
= 2*10^(-4)
= D

Cheers,
Brent
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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11 Apr 2016, 06:19
7
$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

Hi,

we should be able to take advantage of choices whereever possible...
Ofcourse, I donot think choices here were given to be able to eliminate all except ONE...
But then that is what is possible in this Q with the choices given...

$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$..
both the terms individually are $$\frac{ODD}{ODD}$$so each term should come out as ODD and $$ODD - ODD =EVEN$$...
$$\frac{0.99999999*1.0003-1.0001*0.99999991}{1.0003*1.0001}=$$ should be $$\frac{EVEN}{ODD}$$..
so our answer should be something with the last digit in DECIMALs as some EVEN number..
Only D has a 2 in its ten-thousandths place..
D

But ofcourse if we had another choice of same type, we would have had to use a^2-b^2 as done by bunuel But if choice permits, GMAT is all about using the opportunities...
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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Updated on: 03 May 2016, 08:19
5
$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

When first looking at this problem, we must consider the fact that 0.99999999/1.0001 and 0.99999991/1.0003 are both pretty nasty-looking fractions. However, this is a situation in which we can use the idea of the difference of two squares to our advantage. To make this idea a little clearer, let’s first illustrate the concept with a few easier whole numbers. For instance, let’s say we were asked:

999,999/1,001 – 9,991/103 = ?

We could rewrite this as:

(1,000,000 – 1)/1,001 – (10,000 – 9)/103

(1000 + 1)(1000 – 1)/1,001 – (100 – 3)(100 + 3)/103

(1,001)(999)/1,001 – (97)(103)/103

999 – 97 = 902

Notice how cleanly the denominators canceled out in this case. Even though the given problem has decimals, we can follow the same approach.

0.99999999/1.0001 – 0.99999991/1.0003

[(1 – 0.00000001)/1.0001] – [(1 – 0.00000009)/1.0003]

[(1 – 0.0001)(1 + 0.0001)/1.0001] – [(1 – 0.0003)(1 + 0.0003)]

When converting this using the difference of squares, we must be very careful not to make any mistakes with the number of decimal places in our values. Since 0.00000001
has 8 decimal places, the decimals in the factors of the numerator of the first set of brackets must each have 4 decimal places. Similarly, since 0.00000009 has 8 decimal places, the decimals in the factors of the numerator of the second set of brackets must each have 4 decimal places. Let’s continue to simplify.

[(1 – 0.0001)(1 + 0.0001)/1.0001] – [(1 – 0.0003)(1 + 0.0003)]

[(0.9999)(1.0001)/1.0001] – [(0.9997)(1.0003)/1.0003]

0.9999 – 0.9997

0.0002

Converting this to scientific notation to match the answer choices, we have:

2 x 10^-4

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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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10 Jun 2017, 20:27
4
1
Here's how I solved it, which I think is less painful (but maybe wouldn't generalize as well)

$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}$$

Get rid of that distracting decimal:

$$\frac{9999999}{100010000}-\frac{99999991}{100030000}$$

Recognize that both fractions are some very small number from 1, so try to expose that small number by pulling out the 1:

$$\frac{100010000-10001}{100010000}-\frac{100030000-30009}{100030000}$$

Simplify, and marvel at the convenient numbers revealed

$$1-\frac{1}{10000}-1+\frac{3}{10000}$$

Do the arithmetic, done.
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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19 Nov 2017, 03:49
3
$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) $$10^{(-8)}$$

(B) $$3*10^{(-8)}$$

(C) $$3*10^{(-4)}$$

(D) $$2*10^{(-4)}$$

(E) $$10^{(-4)}$$

Simplest way to handle this kind of problem is:

.99 can be written as $$\frac{99}{10^2}$$ OR $$\frac{(10^2-1)}{10^2}$$ and .91 can be written as $$\frac{91}{10^2}$$ OR $$\frac{(10^2-9)}{10^2}$$

For two 9's, we wrote $$\frac{(10^2-1)}{10^2}$$, for eight 9's, we will take $$\frac{(10^8-1)}{10^8}$$
Same rule applies to 1.0001 and 1.0003

So, the simplified version becomes

=> $$\frac{(10^8-1)*10^4}{(10^4+1)*10^8}$$ - $$\frac{(10^8-9)*10^4}{(10^4+1)*10^8}$$

=> $$\frac{{(10^4)^2-1^2}}{10^4*(10^4+1)}$$ - $$\frac{{(10^4)^2-3^2}}{10^4*(10^4+3)}$$

=> $$\frac{(10^4+1)*(10^4-1)}{10^4*(10^4+1)}$$ - $$\frac{(10^4+3)*(10^4-3)}{10^4*(10^4+3)}$$

=> $$\frac{(10^4-1)}{10^4}$$ - $$\frac{(10^4-3)}{10^4}$$

=> $$\frac{1}{10^4}*(10^4-1-10^4+3)$$

=> $$2*10^{-4}$$
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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14 Dec 2017, 23:45
1
VeritasPrepKarishma wrote:
$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

Responding to a pm:
To be honest, I can't think of an alternative method. The fractions are really complicated and need to be simplified before proceeding. For simplification, I think you will need to use a^2 - b^2 = (a - b)(a + b)

All I can suggest is that you can try to solve it without the exponents if that seems easier e.g.

$$\frac{0.99999999}{1.0001}-\frac{.99999991}{1.0003}$$

$$\frac{{1 - .00000001}}{{1 + .0001}}-\frac{{1 - .00000009}}{{1 + .0003}}$$

$$\frac{{1^2 - .0001^2}}{{1 + .0001}}-\frac{{1^2 - 0.0003^2}}{{1 + .0003}}$$

$$\frac{{(1 - .0001)(1 + .0001)}}{{(1 + .0001)}}-\frac{{(1 - .0003)(1 + .0003)}}{{(1 + .0003)}}$$

$$(1 - .0001) - (1 - .0003)$$

$$.0002 = 2*10^{-4}$$

Here is the video solution to this difficult looking problem: https://www.veritasprep.com/gmat-soluti ... olving_211
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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06 Jan 2018, 04:27
Bunuel wrote:
$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}$$

Now apply $$a^2-b^2=(a+b)(a-b)$$:

$$\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}=\frac{(1+10^{-4})(1-10^{-4})}{1+10^{-4}}-\frac{(1+3*10^{-4})(1-3*10^{-4})}{1+3*10^{-4}}=(1-10^{-4})-(1-3*10^{-4})=2*10^{-4}$$.

Hello Bunuel - do you have some useful links for better understanding of your solution ... i dont understand based on which rule transform 1.0003 into 1+3*10^-4 , and 1-9*10^-8 ?

Also what made you apply this formula $$a^2-b^2=(a+b)(a-b)$$ and could you you break down this solution into more detaled steps ?
$$\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}=\frac{(1+10^{-4})(1-10^{-4})}{1+10^{-4}}-\frac{(1+3*10^{-4})(1-3*10^{-4})}{1+3*10^{-4}}=(1-10^{-4})-(1-3*10^{-4})=2*10^{-4}$$. and what common denominator you chose after applying $$a^2-b^2=(a+b)(a-b)$$

Thank you
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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06 Jan 2018, 05:53
dave13 wrote:
Bunuel wrote:
$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$$

(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)

$$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}$$

Now apply $$a^2-b^2=(a+b)(a-b)$$:

$$\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}=\frac{(1+10^{-4})(1-10^{-4})}{1+10^{-4}}-\frac{(1+3*10^{-4})(1-3*10^{-4})}{1+3*10^{-4}}=(1-10^{-4})-(1-3*10^{-4})=2*10^{-4}$$.

Hello Bunuel - do you have some useful links for better understanding of your solution ... i dont understand based on which rule transform 1.0003 into 1+3*10^-4 , and 1-9*10^-8 ?

Also what made you apply this formula $$a^2-b^2=(a+b)(a-b)$$ and could you you break down this solution into more detaled steps ?
$$\frac{1-10^{-8}}{1+10^{-4}}-\frac{1-9*10^{-8}}{1+3*10^{-4}}=\frac{(1+10^{-4})(1-10^{-4})}{1+10^{-4}}-\frac{(1+3*10^{-4})(1-3*10^{-4})}{1+3*10^{-4}}=(1-10^{-4})-(1-3*10^{-4})=2*10^{-4}$$. and what common denominator you chose after applying $$a^2-b^2=(a+b)(a-b)$$

Thank you

1.
$$1+10^{-4}=1+\frac{1}{1,0000}=1+0.0001=1.0001$$
$$1+3*10^{-4}=1+\frac{3}{1,0000}=1+0.0003=1.0003$$

2. You should apply $$a^2-b^2=(a+b)(a-b)$$ because after this you can reduce the denominator.

3.

8. Exponents and Roots of Numbers

Check below for more:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =  [#permalink]

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25 Feb 2018, 16:25
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Re: 0.99999999/1.0001 - 0.99999991/1.0003 =   [#permalink] 25 Feb 2018, 16:25

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