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

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Manager
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If P and Q are positive integers, and n is the decimal  [#permalink]

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Updated on: 18 Nov 2012, 04:50
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Difficulty:

55% (hard)

Question Stats:

60% (01:40) correct 40% (01:57) wrong based on 453 sessions

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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

Originally posted by Jp27 on 17 Nov 2012, 09:48.
Last edited by Bunuel on 18 Nov 2012, 04:50, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If P and Q are positive integers,  [#permalink]

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17 Nov 2012, 09:51
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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Re: If P and Q are positive integers,  [#permalink]

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17 Nov 2012, 11:57
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.
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Re: If P and Q are positive integers, and n is the decimal  [#permalink]

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18 Nov 2012, 13:49
1
It is given in the question stem that P and Q are positive integers.
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Re: If P and Q are positive integers, and n is the decimal  [#permalink]

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20 Jul 2013, 10:51
1
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)
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Re: If P and Q are positive integers,  [#permalink]

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20 Jul 2013, 23:08
1
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...
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Re: If P and Q are positive integers, and n is the decimal  [#permalink]

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20 Jul 2013, 23:49
1
1
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
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Re: If P and Q are positive integers,  [#permalink]

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20 Jul 2013, 23:57
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...

I dont think so... In 121/81 , 81 can be expressed as powers of prime factor(3), but the fraction will not be finite
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Re: If P and Q are positive integers,  [#permalink]

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20 Jul 2013, 23:59
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 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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Re: If P and Q are positive integers,  [#permalink]

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21 Jul 2013, 01:42
1
1
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.

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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Re: If P and Q are positive integers, and n is the decimal  [#permalink]

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11 Mar 2015, 13:53
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).
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Re: If P and Q are positive integers, and n is the decimal  [#permalink]

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31 Dec 2018, 01:37
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?
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Re: If P and Q are positive integers, and n is the decimal  [#permalink]

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06 Sep 2019, 05:16
narendran1990 wrote:
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
Re: If P and Q are positive integers, and n is the decimal   [#permalink] 06 Sep 2019, 05:16
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