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# If each of the following fractions were written as a repeati

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If each of the following fractions were written as a repeati [#permalink]  28 Jan 2008, 15:18
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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
[Reveal] Spoiler: OA
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Re: PS - Number properties [#permalink]  28 Jan 2008, 16:50
I cant think of anything shorter. But, the way I do it is
for 2/11. I make it 20/11. So the quotient will be 1.8181..Then I shift the decimal so it is 0.18

10/3 = 3.333 = 0.3
410/99 = 4.11 = 0.41
20/3 = 6.666 = 0.6
230/37 = 6.21621 = 0.621

So 23/37 has the longest sequence of digits
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Re: PS - Number properties [#permalink]  28 Jan 2008, 18:39
gmatraider wrote:
Is there any easy way to decipher this question (i.e. how do you get the answer here without a calculator or doing long division for a while?)?

If each of the following fractions were written as a repeating decimal, which would have the longest sequence of digits?

2/11
1/3
41/99
2/3
23/37

I also waste more than 2 minutes for this. At first, I did not know what the repeating decimal, only decimal. Anyway, let call to Walker, Alo Walker, where are you? and how are you? hihi
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Re: PS - Number properties [#permalink]  28 Jan 2008, 22:07
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E

After reading of the problem I pick E by intuition: I thought that:
1. nominators do not influence on length of sequence. (for right fraction)
2.the largest prime in denominator lead to the largest sequence).

But I did not know a general rule...

$$1/a=0.(b)=b*(10^{-n}+10^{-2n}+10^{-3n.....}+...)=b*\frac{10^{-n}}{1-10^{-n}}=b*\frac{1}{10^{n}-1}=\frac{b}{99...99_n}$$
where n - the length of sequence (the number of digits of b)

Therefore, $$a$$ have to be a factor of $$99...9_n$$

1. n=1, 9: B,D out, because of 9=3*k
2. n=2, 99: A,C out, because of 99=11*k=99*k

E remains.
It took about 2 min but I think we have now a good rule

Check:

2/11=0.(18) n=2. ok
1/3=0.(3) n=1. ok
41/99=0.(41) n=2. ok
1/3=0.(6) n=1. ok
23/37=0.(621) n=3. 999=27*37. ok
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Re: PS - Number properties [#permalink]  29 Jan 2008, 00:01
walker wrote:
E

After reading of the problem I pick E by intuition: I thought that:
1. nominators do not influence on length of sequence. (for right fraction)
2.the largest prime in denominator lead to the largest sequence).

But I did not know a general rule...

$$1/a=0.(b)=b*(10^{-n}+10^{-2n}+10^{-3n.....}+...)=b*\frac{10^{-n}}{1-10^{-n}}=b*\frac{1}{10^{n}-1}=\frac{b}{99...99_n}$$
where n - the length of sequence (the number of digits of b)

Therefore, $$a$$ have to be a factor of $$99...9_n$$

1. n=1, 9: B,D out, because of 9=3*k
2. n=2, 99: A,C out, because of 99=11*k=99*k

E remains.
It took about 2 min but I think we have now a good rule

Check:

2/11=0.(18) n=2. ok
1/3=0.(3) n=1. ok
41/99=0.(41) n=2. ok
1/3=0.(6) n=1. ok
23/37=0.(621) n=3. 999=27*37. ok

I got it, Walker!

Let apply it, if something wrong, please correct me!
-ignore the denominator as not prime because the largest prime in denominator, the largest sequence
-[Most important] is to check what denominator as prime biggest? right?

Walker, you are deserved to be 51 in Q! Many thanks
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Re: PS - Number properties [#permalink]  29 Jan 2008, 01:27
Expert's post
walker wrote:
After reading of the problem I pick E by intuition: I thought that:
1. nominators do not influence on length of sequence. (for right fraction) - it is true
2.the largest prime in denominator lead to the largest sequence). - it is false

I think a fast way for such problem is:

1. Test for right fractions.
2. Exclude nominators
3. Exclude 2 and 5 from a denominators.
4. 3,9 - n=1, 3,11,33,99 - n=2
5. POI
6. find 1/denominator for remained variants.

For example,
A.3/22
B.5/198

1. ok
2. 1/22, 1/198
3. 1/11, 1/99
4. n=2, n=2
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Re: PS - Number properties [#permalink]  29 Jan 2008, 01:55
walker wrote:
walker wrote:
After reading of the problem I pick E by intuition: I thought that:
1. nominators do not influence on length of sequence. (for right fraction) - it is true
2.the largest prime in denominator lead to the largest sequence). - it is false

I think a fast way for such problem is:

1. Test for right fractions.
2. Exclude nominators
3. Exclude 2 and 5 from a denominators.
4. 3,9 - n=1, 3,11,33,99 - n=2
5. POI= what abbreviation?
6. find 1/denominator for remained variants. Please do not tell me that I must do a calculation!

For example,
A.3/22
B.5/198

1. ok
2. 1/22, 1/198
3. 1/11, 1/99
4. n=2, n=2

Walker, 5 and 6, what is it?
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Re: PS - Number properties [#permalink]  29 Jan 2008, 02:14
Ur basic calculations shiould be strong enough.keep all the basic divisions on ur tips to deal with any big calculations.
Eg.3/22 could be calculated as simple as it is done here.
1/11=0.0909..
3/11 is trice this i.e 0.272727...
0272727..../2 gives u the requiered value..
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Re: PS - Number properties [#permalink]  29 Jan 2008, 02:16
Expert's post
Sorry, P.O.E. (process of elimination)
at 6. You may have some options that did not eliminated by previous steps: for example, 1/7 or 1/37.
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Re: PS - Number properties [#permalink]  29 Jan 2008, 02:22
Expert's post
Pzazz wrote:
Ur basic calculations shiould be strong enough.keep all the basic divisions on ur tips to deal with any big calculations.
Eg.3/22 could be calculated as simple as it is done here.
1/11=0.0909..
3/11 is trice this i.e 0.272727...
0272727..../2 gives u the requiered value..

You are right. I think it is better to have 2-3 ways to solve the problem. You always can choose the most clearest and fastest way that is appropriate for you.
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Re: PS - Number properties [#permalink]  29 Jan 2008, 12:11
gmatraider wrote:
Is there any easy way to decipher this question (i.e. how do you get the answer here without a calculator or doing long division for a while?)?

If each of the following fractions were written as a repeating decimal, which would have the longest sequence of digits?

2/11
1/3
41/99
2/3
23/37

Well I dunno a fast approach other than long division. But you can easily elim B and D b/c they are essentially the same type of repeating decimal.

2/11 --> .1818
41/99 --> .414141 etc...

23/37 --> .621 winner. E it is.

You dont have to go all the way w/ long division here Ex/ 37)23 --> 37)230 --> 6*37 230-222 --> 8 37 goes into 80 twice so 80-74 =6 You should realize here that you will have a different number so this is the answer.

You can apply this approach to the other two as well.
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Re: PS - repeating decimal [#permalink]  05 Apr 2008, 01:25
AlbertNTN wrote:
If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?
2/11
1/3
41/99
2/3
23/37
I am marching thru the OG, this is simple, but just wonder anyone got better trick for repeating decimal (hmm obviously if we play division game for A, B, D we can find; but i try to optimize calculation time )

I remember in one post Walker had some nice explaination about the terminating and repeating fraction.

He said that "1/17 has how many repeating numbers? (p-1) = 16 numbers". So I think E is my choice because E will have 37-1 =36 repeating digits?

Correct me if I wrong you and Walker
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Re: PS - repeating decimal [#permalink]  05 Apr 2008, 01:25
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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.
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Re: PS - repeating decimal [#permalink]  05 Apr 2008, 01:28
Expert's post
sondenso wrote:
I remember in one post Walker had some nice explaination about the terminating and repeating fraction.

He said that "1/17 has how many repeating numbers? (p-1) = 16 numbers". So I think E is my choice because E will have 37-1 =36 repeating digits?

Correct me if I wrong you and Walker

I remember that I even proved rule with 99....999 by means of geometrical progression.... But it say that we should count 9 in 99...999 denominator.
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Re: PS - repeating decimal [#permalink]  05 Apr 2008, 04:27
nice to see the 9 rule, is it some fundamental property of arithmetic? thanks
sodenso i don't think the rule works with 1/37 (only 3 repeated numbers) ...
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Re: PS - repeating decimal [#permalink]  05 Apr 2008, 07:03
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I found that:

7-t59213

7-t57168
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Re: PS - repeating decimal [#permalink]  01 May 2008, 04:32
AlbertNTN wrote:
If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?
2/11
1/3
41/99
2/3
23/37
I am marching thru the OG, this is simple, but just wonder anyone got better trick for repeating decimal (hmm obviously if we play division game for A, B, D we can find; but i try to optimize calculation time )

I did this in my head in bout 2 min. its not really that bad. B,D elim right off the bat.

A. just solve out, you get .1818
C again just solve out you get .4141

No need to test for E.
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Tips and Tricks [#permalink]  28 Jul 2009, 04:10
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I was wondering if there are any tips or tricks to save time in questions like these rather than manually calculating values. I know one trick which is used to check the greater values in fractions

For e.g to check which one greater among say 2/3 , 3/16 and 7/ 48

Just multiply the denominator one fraction to other numerator i.e 2 x 16 and 3 x 3. here 2x16 is greater hence the 2/3 is greater than 3/16.

Folks please share if you know tips and tricks to save time while doing these cheesy time consuming calculations.

Thanks
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Re: Tips and Tricks [#permalink]  28 Jul 2009, 12:16
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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

Last edited by coelholds on 28 Jul 2009, 13:55, edited 2 times in total.
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Re: Tips and Tricks [#permalink]  29 Jul 2009, 05:25
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You are welcome and thanks
Re: Tips and Tricks   [#permalink] 29 Jul 2009, 05:25

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