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vaivish1723
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Does the decimal equivalent of p/q, where p and q are positive integer Updated on: 15 Sep 2020, 07:29
Originally posted by vaivish1723 on 23 Jan 2010, 00:37.
Last edited by Bunuel on 15 Sep 2020, 07:29, edited 3 times in total.
Renamed the topic and edited the question.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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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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Does the decimal equivalent of p/q, where p and q are positive integer 23 Jan 2010, 00:54
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vaivish1723
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
Show 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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Does the decimal equivalent of p/q, where p and q are positive integer 18 Feb 2010, 07:43
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nickk
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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Does the decimal equivalent of p/q, where p and q are positive integer 29 Jan 2010, 09:22
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Bunuel
vaivish1723
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
Show 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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Does the decimal equivalent of p/q, where p and q are positive integer 29 Jan 2010, 10:18
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nverma
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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Does the decimal equivalent of p/q, where p and q are positive integer 18 Feb 2010, 05:29
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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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Does the decimal equivalent of p/q, where p and q are positive integer 16 Sep 2014, 19:01
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Bunuel
vaivish1723
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
Show 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 positive integer 17 Sep 2014, 00:15
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russ9
Bunuel
vaivish1723
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
Show 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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Does the decimal equivalent of p/q, where p and q are positive integer 25 Dec 2014, 12:30
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Bunuel
vaivish1723
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
Show 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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Does the decimal equivalent of p/q, where p and q are positive integer 25 Dec 2014, 12:47
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Ralphcuisak
Bunuel
vaivish1723
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
Show 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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Does the decimal equivalent of p/q, where p and q are positive integer 02 May 2018, 10:37
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vaivish1723
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
Show 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.



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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Bunuel
vaivish1723
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
Show 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.



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 2-s and/or 5-s 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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Does the decimal equivalent of p/q, where p and q are positive integer 29 Jul 2019, 14:29
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Thank you, Bunuel. Your explanations are simply awesome..
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Does the decimal equivalent of p/q, where p and q are positive integer 10 Mar 2020, 02:14
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Bunuel

The fraction 10/7 has 15 digits after decimal but then it terminates. Doesn't it make 10/7 a terminating decimal even though the denominator is 7?
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Does the decimal equivalent of p/q, where p and q are positive integer 10 Mar 2020, 02:16
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ImranBinkhaled
Bunuel

The fraction 10/7 has 15 digits after decimal but then it terminates. Doesn't it make 10/7 a terminating decimal even though the denominator is 7?

10/7 does NOT terminate: 1.428571428571428571428571428571428571428571428571428571428...
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Does the decimal equivalent of p/q, where p and q are positive integer 25 Apr 2022, 10:44
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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

Dear All,

I do not understand why (2) alone is sufficient. If p>q and p is a multiple of q then for the decimal equivalent of p/q we should have infinite zero digits. Could you please help me?
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Does the decimal equivalent of p/q, where p and q are positive integer 09 Oct 2022, 13:37
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Bunuel
vaivish1723
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
Show 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.



Can you please tell me why they are telling us P and Q are positive here?
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lnyngayan
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Does the decimal equivalent of p/q, where p and q are positive integer 21 Sep 2024, 20:48
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Bunuel
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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


Please explain

Oa is
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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.

So for \(2^n5^m\), it could be \(5^0 or 2^0\)?
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Does the decimal equivalent of p/q, where p and q are positive integer 22 Sep 2024, 01:12
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lnyngayan
Bunuel
vaivish1723
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
Show 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.

So for \(2^n5^m\), it could be \(5^0 or 2^0\)?
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