If P and Q are positive integers, and n is the decimal

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

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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.

It is given in the question stem that P and Q are positive integers.

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.

Correct answer: E (I, II, and III)

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

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

I dont think so... In 121/81 , 81 can be expressed as powers of prime factor(3), but the fraction will not be finite

That is not entirely correct. What you state is valid only for 2,5 or both.Also, the given fraction should be a reduced fraction.
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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.

Questions testing this concept:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html

Hope it helps.
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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).
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I can understand that option 1 & 2 's denominator can be expressed in terms of 2^m, but in option 3, 384 cannot be fully expressed in powers of 2. So, how can it be terminating decimal?

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

Theory to Question:
Conditions for a fraction to have terminating decimal:
i. Denominator can be reduced to 2 (power m) 5 (power n)
ii. Denominator has only 2-s and/or 5-s as prime number

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

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.

Questions testing this concept:
https://gmatclub.com/forum/does-the-dec ... 89566.html
https://gmatclub.com/forum/any-decimal- ... 01964.html
https://gmatclub.com/forum/if-a-b-c-d-a ... 25789.html
https://gmatclub.com/forum/700-question-94641.html
https://gmatclub.com/forum/is-r-s2-is-a ... 91360.html
https://gmatclub.com/forum/pl-explain-89566.html
https://gmatclub.com/forum/which-of-the ... 88937.html

Hope it helps.[/quote]

Just to clarify does the fraction need to have either 2 or 5 (or both) ONLY in its denominator or it can have other primes in there as well e.g. 2 * 7 or 5*11

If the fraction is already reduced to its lowest form, it should have only 2s, only 5s, or only 2s and 5s to be a terminating decimal.
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