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

Question

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

Marcab · Accepted Answer

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

jjack0310 · Answer

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)

Jp27 · Answer

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

MacFauz · Answer

]
 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.

watwazdaquestion · Answer

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

avinashrao9 · Answer

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

swati007 · Answer

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

mau5 · Answer

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.

Bunuel · Answer

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.

thefibonacci · Answer

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).

narendran1990 · Answer

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?

thaya · Answer

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

ConfusedSEA · Answer

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

Syavashp · Answer

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. 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. 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

Bunuel · Answer

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.

bumpbot · Answer

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

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

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