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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.
_________________

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.

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.

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.

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?

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

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06 May 2014, 19:35

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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?

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.
_________________

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?

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.
_________________

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

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18 May 2016, 02:19

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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