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Does the decimal equivalent of P/Q, where P and Q are
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Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? (1) P>Q (2) Q=8
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Originally posted by vaivish1723 on 23 Jan 2010, 00:37.
Last edited by Bunuel on 23 May 2013, 05:03, edited 2 times in total.
Edited the OA, it must be B not E




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Does the decimal equivalent of P/Q, where P and Q are
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23 Jan 2010, 00:54
vaivish1723 wrote: 26 Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? (1) P>Q (2) Q=8 Please explain Oa is Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Question: Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? According to the above we must determine whether the denominator (after reducing the fraction, if possible) contains only the 2s and/or 5s as the prime factors. (1) P>Q, clearly insufficient. (2) Q=8=2^3, hence denominator has only 2 as prime factor. Fraction P/Q will be terminated decimal. Sufficient. Answer: B.
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Re: pl explain
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29 Jan 2010, 09:22
Bunuel wrote: vaivish1723 wrote: 26 Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? (1) P>Q (2) Q=8 Please explain Oa is Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Question: Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? According to the above we must determine whether the denominator (after reducing the fraction, if possible) contains only the 2s and/or 5s as the prime factors. (1) P>Q, clearly insufficient. (2) Q=8=2^3, hence denominator has only 2 as prime factor. Fraction P/Q will be terminated decimal. Sufficient. Answer: B. (OA must be wrong) Hello Bunuel How can we say that the fraction given is a "reduced fraction". Because if it's not than 70/8 is a nonterminating value.



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29 Jan 2010, 10:18



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18 Feb 2010, 05:29
Bunuel please excuse the stupid question but I'm quite weak in these types of questions.
* Does the rule basically say that any integer divided by either 2, 5, a multiple of either, or a product of these multiples will have a finite amount of decimals?
So any integer divided by any multiple of 5 will have a finite amount of decimals, etc.?
* And another basic question: 0 is neither positive nor negative right?
* I also don't completely understand what to do when we don't know whether the fraction is reduced to its lowest form. Can we still apply the same rule?



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18 Feb 2010, 07:43
nickk wrote: Bunuel please excuse the stupid question but I'm quite weak in these types of questions.
1.Does the rule basically say that any integer divided by either 2, 5, a multiple of either, or a product of these multiples will have a finite amount of decimals?
2. So any integer divided by any multiple of 5 will have a finite amount of decimals, etc.?
3. And another basic question: 0 is neither positive nor negative right?
4. I also don't completely understand what to do when we don't know whether the fraction is reduced to its lowest form. Can we still apply the same rule?
1. Not multiples but when denominator has only 2 and/or 5 in any integer power; 2. No, multiples of 5 are 5, 10, 15, 20, 25, 30, ... 1/30 won't be terminating as there is 3 in denominator. Maybe you meant 5 in any power? Then yes; 3. Yes, (though it's even); 4. If fraction has only 2 or/and 5 in denominator then it does not matter whether it's reduced. If there is some other integer in denominator we need reducing to see whether it can be cancelled.
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18 Feb 2010, 08:01
Thanks! that explains everything



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06 Oct 2011, 07:34
How to know these properties before, some book/site has these properties covered thoroughly.



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06 Oct 2011, 07:41
arnoorichenna wrote: How to know these properties before, some book/site has these properties covered thoroughly. gmatmathbook87417.htmlMGMAT Strategy Guides 4th Edition OG12, OG Quantitative Review 2 Nothing else is required beyond these for GMAT related quantitative concepts.
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06 Oct 2011, 13:42
It is a repeat soln is explained in 700question94641.html
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Re: Does the decimal equivalent of P/Q, where P and Q are
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16 Sep 2014, 19:01
Bunuel wrote: vaivish1723 wrote: 26 Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? (1) P>Q (2) Q=8 Please explain Oa is Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Question: Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? According to the above we must determine whether the denominator (after reducing the fraction, if possible) contains only the 2s and/or 5s as the prime factors. (1) P>Q, clearly insufficient. (2) Q=8=2^3, hence denominator has only 2 as prime factor. Fraction P/Q will be terminated decimal. Sufficient. Answer: B. Hi Bunuel, One question  if we know that the denominator only has powers of 5 OR only has powers of 2  it's irrelevant if the denominator is great or less than the numerator, the fraction will still end up in a terminating decimal. is that correct?



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Does the decimal equivalent of P/Q, where P and Q are
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17 Sep 2014, 00:15
russ9 wrote: Bunuel wrote: vaivish1723 wrote: 26 Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? (1) P>Q (2) Q=8 Please explain Oa is Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Question: Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? According to the above we must determine whether the denominator (after reducing the fraction, if possible) contains only the 2s and/or 5s as the prime factors. (1) P>Q, clearly insufficient. (2) Q=8=2^3, hence denominator has only 2 as prime factor. Fraction P/Q will be terminated decimal. Sufficient. Answer: B. Hi Bunuel, One question  if we know that the denominator only has powers of 5 OR only has powers of 2  it's irrelevant if the denominator is great or less than the numerator, the fraction will still end up in a terminating decimal. is that correct? Yes. \(\frac{integer}{2^n5^m}\), for nonnegative integers n and m, will be a terminating decimal irrespective of the numerator.
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Re: Does the decimal equivalent of P/Q, where P and Q are
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25 Dec 2014, 12:30
Bunuel wrote: vaivish1723 wrote: 26 Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? (1) P>Q (2) Q=8 Please explain Oa is Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Question: Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? According to the above we must determine whether the denominator (after reducing the fraction, if possible) contains only the 2s and/or 5s as the prime factors. (1) P>Q, clearly insufficient. (2) Q=8=2^3, hence denominator has only 2 as prime factor. Fraction P/Q will be terminated decimal. Sufficient. Answer: B. the doubt i have regarding this concept is: The prime factors of the denominator cannot be anything except 2 and/or 5? eg. 1/30 will not be a terminating decimal as it has 3 too?



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Re: Does the decimal equivalent of P/Q, where P and Q are
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25 Dec 2014, 12:47
Ralphcuisak wrote: Bunuel wrote: vaivish1723 wrote: 26 Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? (1) P>Q (2) Q=8 Please explain Oa is Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Question: Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? According to the above we must determine whether the denominator (after reducing the fraction, if possible) contains only the 2s and/or 5s as the prime factors. (1) P>Q, clearly insufficient. (2) Q=8=2^3, hence denominator has only 2 as prime factor. Fraction P/Q will be terminated decimal. Sufficient. Answer: B. the doubt i have regarding this concept is: The prime factors of the denominator cannot be anything except 2 and/or 5? eg. 1/30 will not be a terminating decimal as it has 3 too? Yes, 1/30 will not be a terminating decimal.
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Re: Does the decimal equivalent of P/Q, where P and Q are
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02 May 2018, 10:37
Bunuel wrote: vaivish1723 wrote: 26 Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? (1) P>Q (2) Q=8 Please explain Oa is Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Question: Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? According to the above we must determine whether the denominator (after reducing the fraction, if possible) contains only the 2s and/or 5s as the prime factors. (1) P>Q, clearly insufficient. (2) Q=8=2^3, hence denominator has only 2 as prime factor. Fraction P/Q will be terminated decimal. Sufficient. Answer: B. Hi Bunuel, In this question it was not mentioned that P/Q is in reduced form So how can we use the 2\sqrt{m}*5\sqrt{n}. Is it assumed that it is in reduced form?? Thanks in advance



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Re: Does the decimal equivalent of P/Q, where P and Q are
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02 May 2018, 12:21
suramya26 wrote: Bunuel wrote: vaivish1723 wrote: 26 Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? (1) P>Q (2) Q=8 Please explain Oa is Theory:Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\). Question: Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits? According to the above we must determine whether the denominator (after reducing the fraction, if possible) contains only the 2s and/or 5s as the prime factors. (1) P>Q, clearly insufficient. (2) Q=8=2^3, hence denominator has only 2 as prime factor. Fraction P/Q will be terminated decimal. Sufficient. Answer: B. Hi Bunuel, In this question it was not mentioned that P/Q is in reduced form So how can we use the 2\sqrt{m}*5\sqrt{n}. Is it assumed that it is in reduced form?? Thanks in advance If denominator already has only 2s and/or 5s then it doesn't matter whether the fraction is reduced or not. For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal. We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.
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Re: Does the decimal equivalent of P/Q, where P and Q are
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18 Jun 2018, 03:07
vaivish1723 wrote: Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits?
(1) P>Q (2) Q=8 What does this question mean?



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Re: Does the decimal equivalent of P/Q, where P and Q are
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18 Jun 2018, 22:22
Mahadi Hasan wrote: vaivish1723 wrote: Does the decimal equivalent of P/Q, where P and Q are positive integers, contain only a finite number of nonzero digits?
(1) P>Q (2) Q=8 What does this question mean? Lets look at a fraction 1/2. When you convert 1/2 to decimal, what do you get: 0.5. This has a finite number of nonzero digits (only one non zero digit, which is '5'). Now consider fraction 1/3. When you convert 1/3 to decimal, you get 0.333333... (unending). This contains infinite number of nonzero digits. So the question is asking whether P/Q has a finite or an infinite number of nonzero digits?




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