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Re: If P and Q are positive integers, and n is the decimal [#permalink]
20 Jul 2013, 09:51

A. P/Q = (49/256)= (7)(7)/(16)(16) P/Q = (7/16)*(7/16) = .4125 * .4125 = finite B. 32 = 2^5. Any number (odd/even) divided by 2^n will always be finite. C. 75/384 = (3*5^2)/(2^7*3) ---> 3 gets cancelled and we have 5^2 / 2^7 - always finite because of 2^7.

Re: If P and Q are positive integers, and n is the decimal [#permalink]
20 Jul 2013, 22:49

Expert's post

The best trick to find out whether a fraction will yield a definite decimal number is to check whether the denominator can be expressed in terms of the power of 2 and/or 5. If yes, then the fraction will yield a definite decimal. In the above question 256, 32 and 384 can be expressed in powers of 2 as well. Hence I, II and III are correct. Regards _________________

Re: If P and Q are positive integers, [#permalink]
21 Jul 2013, 00:42

Expert's post

avinashrao9 wrote:

MacFauz wrote:

Jp27 wrote:

My doubt is if it were given P and Q to be positive numbers and I)& III) are only correct right? As the P can be 1/3.

Cheers

] I should think so... Infact.. If it had been given as postive numbers, P could be any irrational number such as \(\sqrt{2},\sqrt{3}, \sqrt{5}\)

So, the answer would be only 1 & 3.

Kudos Please... If my post helped.

As long as the denominator can be expressed as powers of prime factors, the fraction will always be finite...

That's not true. Any positive integer can be expressed as powers of primes.

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 \(10=2*5\).

Note that 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.

Re: If P and Q are positive integers, and n is the decimal [#permalink]
11 Mar 2015, 10:24

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).

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Re: If P and Q are positive integers, and n is the decimal [#permalink]
11 Mar 2015, 12:53

Jp27 wrote:

If P and Q are positive integers, and n is the decimal equivalent of P/Q, which of the following must make n a finite number?

I. P = 49, Q = 256 II. Q = 32 III. P = 75, Q = 384

A. None B. I only C. II only D. III only E. I, II, III

the thing to know here is that in any base x a fraction 1/n (in the smallest form) results in a finite decimal form if n can be represented in power of x or of x's factor(s). _________________

Illegitimi non carborundum.

gmatclubot

Re: If P and Q are positive integers, and n is the decimal
[#permalink]
11 Mar 2015, 12:53

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