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repeating decimals [#permalink]
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10 Feb 2009, 23:19
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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
is there a short cut for this?? seems like this should be a 30 second question



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Re: repeating decimals [#permalink]
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10 Feb 2009, 23:38
bigtreezl 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
is there a short cut for this?? seems like this should be a 30 second question You can easily elimite B & D Than long division for A, C and E to come up with E.



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Re: repeating decimals [#permalink]
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11 Feb 2009, 01:00
You can solve this pretty quickly using approximation.
A) 2/11 can be approx. to 2/10 = 1/5 = 0.2 B) 1/3 = 0.33 C) 41/99 can be approx. to 40/100 = 2/5 = 0.4 D) 2/3 = 0.66 E) 23/37... mmh, the weird one; if you want to double check you can proceed with the division (which yields 0.621, hence the longest sequence of different digits).
Hope this helps!



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Re: repeating decimals [#permalink]
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11 Feb 2009, 01:06
thanks! looking for ways to shave time



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Re: repeating decimals [#permalink]
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11 Feb 2009, 03:51
There is a general rule: 0.(abcd)=abcd/9999. So, the more "9" we need to write out our fraction in such way, the longer sequence we have. (You can use search to find my prove of this rule). Moreover, if we have proper fraction x/y, x doesn't influence on length of a sequence. 1) First of all, let's take a fast look at our options. Are their proper fraction? Or do we need to change them? All fractions are proper fractions. 2) cut numerators. 3) Can we compose 9 out of each denominators? Only for B and D. They are out. 4) Can we compose 99 out of each denominators? Only for A and C. They are out. Only E remains. In other words, 1/37 (23/37) has sequence consisted of more than two digits. If you've truly grasped this way, you can solve problems like that under 20 sec and without complex calculations.
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Re: repeating decimals [#permalink]
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11 Feb 2009, 08:10
walker wrote: There is a general rule: 0.(abcd)=abcd/9999. So, the more "9" we need to write out our fraction in such way, the longer sequence we have. (You can use search to find my prove of this rule). Moreover, if we have proper fraction x/y, x doesn't influence on length of a sequence.
1) First of all, let's take a fast look at our options. Are their proper fraction? Or do we need to change them? All fractions are proper fractions.
2) cut numerators.
3) Can we compose 9 out of each denominators? Only for B and D. They are out.
4) Can we compose 99 out of each denominators? Only for A and C. They are out. Only E remains. In other words, 1/37 (23/37) has sequence consisted of more than two digits.
If you've truly grasped this way, you can solve problems like than under 20 sec and without complex calculations. Walker = Genius +1 Thanks buddy for a great tip.
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Re: repeating decimals [#permalink]
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11 Feb 2009, 08:41
There is a prove. It is better to remember a formula when it is understandable \(1/a=0.(b)=b*(10^{n}+10^{2n}+10^{3n.....}+...)=b*\frac{10^{n}}{110^{n}}=b*\frac{1}{10^{n}1}=\frac{b}{99...99_n}\) where n  the length of sequence (the number of digits of b)
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Re: repeating decimals [#permalink]
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11 Feb 2009, 10:01
walker, how can u b soooooo perfect in maths?Give us also sum tips. U r simply marvellous! needless 2 say +1 frm me walker wrote: There is a general rule: 0.(abcd)=abcd/9999. So, the more "9" we need to write out our fraction in such way, the longer sequence we have. (You can use search to find my prove of this rule). Moreover, if we have proper fraction x/y, x doesn't influence on length of a sequence.
1) First of all, let's take a fast look at our options. Are their proper fraction? Or do we need to change them? All fractions are proper fractions.
2) cut numerators.
3) Can we compose 9 out of each denominators? Only for B and D. They are out.
4) Can we compose 99 out of each denominators? Only for A and C. They are out. Only E remains. In other words, 1/37 (23/37) has sequence consisted of more than two digits.
If you've truly grasped this way, you can solve problems like than under 20 sec and without complex calculations.



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Re: repeating decimals [#permalink]
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11 Feb 2009, 11:10
walker wrote: There is a general rule: 0.(abcd)=abcd/9999. So, the more "9" we need to write out our fraction in such way, the longer sequence we have. (You can use search to find my prove of this rule). Moreover, if we have proper fraction x/y, x doesn't influence on length of a sequence.
1) First of all, let's take a fast look at our options. Are their proper fraction? Or do we need to change them? All fractions are proper fractions.
2) cut numerators.
3) Can we compose 9 out of each denominators? Only for B and D. They are out.
4) Can we compose 99 out of each denominators? Only for A and C. They are out. Only E remains. In other words, 1/37 (23/37) has sequence consisted of more than two digits.
If you've truly grasped this way, you can solve problems like than under 20 sec and without complex calculations. wow this is EXACTLY what I was looking for..gmatclub is blessed to have you math gurus here...thanks alot!



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Re: repeating decimals [#permalink]
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11 Feb 2009, 11:59
walker wrote: There is a general rule: 0.(abcd)=abcd/9999. So, the more "9" we need to write out our fraction in such way, the longer sequence we have. (You can use search to find my prove of this rule). Moreover, if we have proper fraction x/y, x doesn't influence on length of a sequence.
1) First of all, let's take a fast look at our options. Are their proper fraction? Or do we need to change them? All fractions are proper fractions.
2) cut numerators.
3) Can we compose 9 out of each denominators? Only for B and D. They are out.
4) Can we compose 99 out of each denominators? Only for A and C. They are out. Only E remains. In other words, 1/37 (23/37) has sequence consisted of more than two digits.
If you've truly grasped this way, you can solve problems like than under 20 sec and without complex calculations. Words are less to appreciate this explanation. wonderful GMAT Tip. +1 from me too.



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Re: repeating decimals [#permalink]
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11 Feb 2009, 13:29
Maybe once I will write own math guide to GMAT.... who knows....
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Re: repeating decimals [#permalink]
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11 Feb 2009, 13:33
walker wrote: Maybe once I will write own math guide to GMAT.... who knows.... If I am still fighting to meet my target GMAT, then I will definitely get the book.



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Re: repeating decimals [#permalink]
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11 Feb 2009, 13:59
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Re: repeating decimals [#permalink]
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11 Feb 2009, 14:01
walker wrote: :) I am serious.. Why don't you start writing the book.. It will sell like hot cake..
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Re: repeating decimals [#permalink]
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11 Feb 2009, 14:02
walker wrote: Maybe once I will write own math guide to GMAT.... who knows.... you should seriously consider it $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$



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Re: repeating decimals [#permalink]
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11 Feb 2009, 14:10
Hm...I'll think about that crazy idea
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Re: repeating decimals [#permalink]
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11 Feb 2009, 21:23
Genrous walker give his maths book 2 all gmatclubbers for free walker wrote: Hm...I'll think about that crazy idea



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Re: repeating decimals [#permalink]
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07 Dec 2009, 08:18
Walker, that was brilliant! It seems that you know tip for each math problem! How was your GMAT score?



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Re: repeating decimals [#permalink]
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07 Dec 2009, 08:21
Lindsy wrote: Walker, that was brilliant! It seems that you know tip for each math problem! How was your GMAT score? See my data under avatar: 750 Q50V40. I think a few silly mistakes + my primary focus on Verbal cost me 1 point in math.
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Re: repeating decimals [#permalink]
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07 Dec 2009, 13:46
sorry i havent explore your avatar so exactly(( But i have read some of your posts) Amazing for me was that you come from Kiev. I am from Kiev too)




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