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If each of the following fractions were written as a repeati

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Re: "Sometimes you just have to guess" - What Level is this?  [#permalink]

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New post 11 Jul 2011, 23:39
you correctly eliminated B and D.
For rest i think you just have to do long division. Just make sure you do it neatly so as to avoid any chance of silly mistakes.
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Re: "Sometimes you just have to guess" - What Level is this?  [#permalink]

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New post 11 Jul 2011, 23:47
Wouldn't LD take about 1min for each?

I guess I won't know how long it really takes unless I try it and time it...I'm sure it'd take longer than 2mins tho...
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Re: "Sometimes you just have to guess" - What Level is this?  [#permalink]

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New post 12 Jul 2011, 00:21
11
TIP If the denominator is 9, 99, 999 or another number equal to a power of 10 minus 1, then
the numerator gives you the repeating digits
:D
A) 2/11= 18/99 = 0.181818
B) 1/3 you should know that it is 0.333333333
C) 41/99 according to the tip above it should be 0.4141414141
D) 2/3 you should know that it is 0.66666666
E) 23/37 you solve then you see :D
it didn't take more than a minute
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Re: "Sometimes you just have to guess" - What Level is this?  [#permalink]

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New post 18 Jul 2012, 11:16
solved in 00:31

we know that 1/11 = 0.09090...

so we that i was able to eliminate A and C
B and D are easy to eliminate as 1/3 = 0.333

So ans E
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Re: PS - repeating decimal  [#permalink]

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New post 15 May 2014, 12:54
walker wrote:
E

1. reduce fraction if it is possible. Here we have all proper fractions.
2. because numerator does not influence on period of sequence, set all numerators to 1: 1/11, 1/3, 1/99, 1/3, 1/37
3. transform all fraction to the denominator such as 9, 99, 999 .....: 9/99, 3/9, 1/99, 3/9, 1/37
4. 1/37 we cannot write out as a fraction with denominator of 9 or 99. (actually we can with 999 but this is not necessary here) So, E is a winner.


The thing here is that if it has 99 in denominator then it means that the same number repeats over and over again indefinitely, therefore since we are looking for the number with the longest sequence of DIFFERENT DIGITS, then 37 is our best pick

Hope this clarifies
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Re: Tips and Tricks  [#permalink]

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New post 08 Aug 2014, 20:39
coelholds wrote:
I have taken 15 seconds to resolve this problem! For this problem specifically, there are some very fast tricks that you can do remembering the properties of recurring decimals. You just need a little patience to understand a first time. Then you probably will never forget.

Here is a brief description useful in a some GMAT exercises:

First - We use the number that is repeating and the denominator has as many "9" digits as there are different digits in the block that repeats. e.g.
\(0.555555 = 5/9\)
\(0.13131313 = 13/99\)
\(0.432432432432 = 432/999\)

Second - If the sequence starts to repeat after some zeros, add the same number of zeros in the denominator. e.g.
\(0.005555 = 5/900\)
\(0.013131313 = 13/990\)
\(0.0004324324324... = 432/999000 = 54/124875 = 2/4625\)

Third - Terminating decimals + Repeating Decimals. e.g.
\(2.31555555 = 2.31 + 0.00555555 = 231/100 + 5/900\)
\(0.745454545 = 0.7 + 0.0454545 = 7/10 + 45/990\)

Last one - The reciprocal of a prime number "p", except 2 and 5, has a repeating sequence of p-1 digits, or a factor of p-1 digits. e.g.
\(1/7 = 0.142857 142857 142857...\) - As you can notice, the sequence has 6 digits = p-1 = 7-1 = 6 digits.
And if you multiply this fraction by a number you will only change the beginning of the sequence. e.g.
\(4/7 = 4*1/7 = 0.57 142857 142857 ...\)

****************
Once you have all these little rules in mind it is very fast to work in this problem.
A) 2/11 -> 11 is a factor of 99, so there are 2 repeating digits.
B) 1/3 -> 3 is a factor of 9, so there is 1 repeating digit.
C) 41/99 -> 99, so there are 2 repeating digits.
D) 2/3 -> 3 is a factor of 9, so there is 1 repeating digit..
E) 23/37 -> As I still don't have one item that is winning in repeating digits, this must be CORRECT

\(E\)

Good studies!

If you enjoyed the post, please consider a kudo!!! ;) I need to practice my verbal skills in the GMATClub tests :)


Hi.

Thanks for the tips. But I guess last one is not correct.
as 1/11, 1/13 .. don't have repeating sequences of 10 and 12 digits respectively.

what do u say?
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Re: If each of the following fractions were written as a repeati  [#permalink]

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New post 22 Nov 2014, 07:12
A,B,C,D give 9 or 99 in denominators meaning that maximal repeating digits are two.
Only E gives more than 99 (999 in this case) because 37 is not factor of it.


E
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Re: If each of the following fractions were written as a repeating decimal  [#permalink]

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New post 22 Jun 2017, 04:23
Hi all

I came across this question on this forum but the discussions were locked. I am not sure about its source, so I selected others and according to me the difficulty level should be 600.

Please give explanations with the correct answer and also can anyone please explain how to easily simplify fractions to decimals and vice-versa ?

Thanks
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If each of the following fractions were written as a repeating decimal  [#permalink]

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New post 22 Jun 2017, 04:27
Shiv2016 wrote:
If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?

(A) 2/11
(B) 1/3
(C) 41/99
(D) 2/3
(E) 23/37


Check the options.

(A) \(\frac{2}{11} = 0.181818\) = has 2 digits repeating itself.

(B) \(\frac{1}{3} = 0.3333\) = has 1 digit repeating itself.

(C) \(\frac{41}{99} = 0.414141\) = has 2 digits repeating itself.

(D) \(\frac{2}{3} = 0.6666\) = has 1 digit repeating itself.

(E) \(\frac{23}{37} = 0.621621\) = has 3 digits repeating itself.

\(\frac{23}{37}\) has Longest sequence of different digits. Answer (E) ...
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Re: If each of the following fractions were written as a repeati  [#permalink]

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New post 22 Jun 2017, 05:53
gmatraider wrote:
If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?

A) 2/11
B) 1/3
C) 41/99
D) 2/3
E) 23/37


2/11 = 0.1818...
1/3 = 0.3333..
41/99 = 0.414141...
2/3= 0.66666..
23/37 = 0.612.....

So Answer is E.
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Re: If each of the following fractions were written as a repeati  [#permalink]

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New post 24 Jun 2017, 07:31
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gmatraider wrote:
If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?

A) 2/11
B) 1/3
C) 41/99
D) 2/3
E) 23/37


If a fraction (in lowest terms) can be written as a repeating decimal and the number of decimal digits repeating is n, then the denominator must be a factor of 10^n - 1.

For example, 1/3 (in choice B) has a denominator of 3, which is a factor of 9 = 10^1 - 1 = 9; thus, 1/3 has 1 decimal digit repeating. Sure enough, 1/3 = 0.333… Another example, 2/11 (in choice A), has a denominator of 11, which is a factor of 99 = 10^2 - 1 = 99; thus, 2/11 has 2 decimal digits repeating. Sure enough, 2/11 = 0.181818…

Using this fact, we can see that 2/3 (in choice D) and 41/99 (in choice C) have 1 and 2 decimal digits repeating, respectively. Thus, the fraction 23/37 has the longest sequence of repeating decimal digits, since 37 is neither a factor of 9 nor a factor of 99. Regardless of the exact number of repeating decimal digits 23/27 has, it will certainly be more than 2.

Answer: E
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Re: If each of the following fractions were written as a repeati  [#permalink]

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Re: If each of the following fractions were written as a repeati   [#permalink] 21 Jul 2020, 10:20

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