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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 [#permalink]

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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
[Reveal] Spoiler: OA

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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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
[Reveal] Spoiler:
e


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 non-negative 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 2-s and/or 5-s 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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New post 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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New post 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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New post 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
[Reveal] Spoiler:
e


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 non-negative 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 2-s and/or 5-s 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 non-terminating value.

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nverma wrote:
Hello Bunuel

How can we say that the fraction given is a "reduced fraction". Because if it's not than 70/8 is a non-terminating value.


Denominator already has only 2-s so in this case it's 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.

Hope it's clear.
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New post 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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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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New post 18 Feb 2010, 08:01
Thanks! that explains everything

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New post 05 Oct 2011, 22:47
Nice explanation there! Answer should be B
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How to know these properties before, some book/site has these properties covered thoroughly.

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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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New post 06 Oct 2011, 13:42
It is a repeat

soln is explained in

700-question-94641.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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New post 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
[Reveal] Spoiler:
e


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 non-negative 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 2-s and/or 5-s 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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New post 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
[Reveal] Spoiler:
e


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 non-negative 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 2-s and/or 5-s 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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New post 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
[Reveal] Spoiler:
e


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 non-negative 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 2-s and/or 5-s 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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New post 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
[Reveal] Spoiler:
e


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 non-negative 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 2-s and/or 5-s 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   [#permalink] 18 May 2016, 02:19

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