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Does the decimal equivalent of P/Q, where P and Q are [#permalink]
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23 Jan 2010, 00:37
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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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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 [#permalink]
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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 [#permalink]
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23 Jan 2010, 11:01
why am I seeing all the questions with wrong OA today or atleast it seemed to be I too got the ans as B



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Re: pl explain [#permalink]
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26 Jan 2010, 09:55
yes OA must be B Please try to post sources as well, so that we can know credibility of the sources
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Re: pl explain [#permalink]
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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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Re: pl explain [#permalink]
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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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Re: pl explain [#permalink]
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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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Re: pl explain [#permalink]
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18 Feb 2010, 08:01
Thanks! that explains everything



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How to know these properties before, some book/site has these properties covered thoroughly.



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Re: pl explain [#permalink]
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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 [#permalink]
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Re: Does the decimal equivalent of P/Q, where P and Q are [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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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