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adkikani
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If x is a positive integer, x^(1/2) an integer? Updated on: 25 Nov 2017, 00:13
Originally posted by adkikani on 24 Nov 2017, 16:40.
Last edited by Bunuel on 25 Nov 2017, 00:13, edited 2 times in total.
Renamed the topic, edited the question and edited the tags.
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If x is a positive integer, is \(\sqrt{x}\) an integer ?

(1) \(\sqrt{36x}\) is an integer.


(2) \(\sqrt{3x+4}\) is an integer
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If x is a positive integer, x^(1/2) an integer? 24 Nov 2017, 19:42
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If x is a positive integer, is \(\sqrt{x}\) an integer

1. \(\sqrt{36x}\) is an integer.


2. \(\sqrt{3x+4}\) is an integer
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is \(\sqrt{x}\) an integer?

1. \(\sqrt{36x}\) is an integer.
\(\sqrt{36x}=\sqrt{6^2x}=6\sqrt{x}\)
Now if \(6\sqrt{x}\) is a n integer and x is a positive integer, \(\sqrt{x}\) has to be an integer
sufficient

2. \(\sqrt{3x+4}\) is an integer
there can be various possiblities....
x = 4, \(\sqrt{3x+4} = \sqrt{3*4+4}\sqrt{16}=4\)
here \(\sqrt{x} =2\) is an integer ...yes
x = 7, \(\sqrt{3x+4} = \sqrt{3*7+4}\sqrt{25}=5\)
here \(\sqrt{5} =2\) is not an integer ...
insuff

A
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If x is a positive integer, x^(1/2) an integer? 24 Nov 2017, 23:20
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chetan2u niks18 Bunuel

Can any of rules be applied universally?

(1) \(\sqrt{ab}\) = \(\sqrt{a}\) * \(\sqrt{b}\)

is it necessary for a and b to be distinct?


(2) \(\sqrt{a+b}\) = \(\sqrt{a}\) + \(\sqrt{b}\)
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If x is a positive integer, x^(1/2) an integer? 25 Nov 2017, 00:44
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If x is a positive integer, is \(\sqrt{x}\) an integer?

The square root of any positive integer is either an integer or an irrational number. So, \(\sqrt{integer}\) cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{17}\), ...). So, as given that \(x\) is a positive integer then \(\sqrt{x}\) is either an integer itself or an irrational number.

(1) \(\sqrt{36x}\) is an integer:

\(\sqrt{36x} = 6\sqrt{x}\). As discussed above, \(\sqrt{x}\) cannot be a fraction, say 1/6 and integer*irrational cannot be an integer. Therefore, for this statement to be true, \(\sqrt{x}\) must be an integer. Sufficient.

(2) \(\sqrt{3x+4}\) is an integer. If x = 4, then the answer is YES but if x = 7, then the answer is NO. Not sufficient.

Answer: A.

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Hope it helps.
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If x is a positive integer, x^(1/2) an integer? 25 Nov 2017, 00:47
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adkikani
chetan2u niks18 Bunuel

Can any of rules be applied universally?

(1) \(\sqrt{ab}\) = \(\sqrt{a}\) * \(\sqrt{b}\)

is it necessary for a and b to be distinct?


(2) \(\sqrt{a+b}\) = \(\sqrt{a}\) + \(\sqrt{b}\)
Show more

(1) Always true: \(\sqrt{ab}\) = \(\sqrt{a}\) * \(\sqrt{b}\)

(2) Generally, \(\sqrt{a+b}\) \neq \(\sqrt{a}\) + \(\sqrt{b}\). Does \(\sqrt{1+1}\)=\(\sqrt{1}\) + \(\sqrt{1}\)?

8. Exponents and Roots of Numbers



Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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If x is a positive integer, x^(1/2) an integer? 25 Nov 2017, 02:38
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adkikani
chetan2u niks18 Bunuel

Can any of rules be applied universally?

(1) \(\sqrt{ab}\) = \(\sqrt{a}\) * \(\sqrt{b}\)

is it necessary for a and b to be distinct?


(2) \(\sqrt{a+b}\) = \(\sqrt{a}\) + \(\sqrt{b}\)
Show more


Hi...
Ofcourse Bunuel has already answered that √a*√b=√(ab)..

Just add on to the second point..
√a+√b will generally be NOT equal to √(a+b)..
Only exception is when either or both of them are 0..
An algebraic way to look at it...

\(√a+√b=√(a+b)\)
Square both sides..
a+b+2√(ab)=a+b........
2√(ab)=0
Only possible When either of a or b or BOTH of a and b are 0..
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If x is a positive integer, x^(1/2) an integer? 06 Jun 2021, 05:35
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chetan2u
adkikani
chetan2u niks18 Bunuel

Can any of rules be applied universally?

(1) \(\sqrt{ab}\) = \(\sqrt{a}\) * \(\sqrt{b}\)

is it necessary for a and b to be distinct?


(2) \(\sqrt{a+b}\) = \(\sqrt{a}\) + \(\sqrt{b}\)


Hi...
Ofcourse Bunuel has already answered that √a*√b=√(ab)..

Just add on to the second point..
√a+√b will generally be NOT equal to √(a+b)..
Only exception is when either or both of them are 0..
An algebraic way to look at it...

\(√a+√b=√(a+b)\)
Square both sides..
a+b+2√(ab)=a+b........
2√(ab)=0
Only possible When either of a or b or BOTH of a and b are 0..
Show more

why cant x be 1/36 then 6* 1/6 = 1 an integer ? @chethan2u
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If x is a positive integer, x^(1/2) an integer? 29 Sep 2021, 04:17
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chetan2u
adkikani
chetan2u niks18 Bunuel

Can any of rules be applied universally?

(1) \(\sqrt{ab}\) = \(\sqrt{a}\) * \(\sqrt{b}\)

is it necessary for a and b to be distinct?


(2) \(\sqrt{a+b}\) = \(\sqrt{a}\) + \(\sqrt{b}\)


Hi...
Ofcourse Bunuel has already answered that √a*√b=√(ab)..

Just add on to the second point..
√a+√b will generally be NOT equal to √(a+b)..
Only exception is when either or both of them are 0..
An algebraic way to look at it...

\(√a+√b=√(a+b)\)
Square both sides..
a+b+2√(ab)=a+b........
2√(ab)=0
Only possible When either of a or b or BOTH of a and b are 0..

why cant x be 1/36 then 6* 1/6 = 1 an integer ? @chethan2u
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We cannot assume \(x = \frac{1}{36}\) because the question mentions that \(x\) is a positive integer and \(\frac{1}{36}\) is not

Hope it clarifies?
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If x is a positive integer, x^(1/2) an integer? 24 Apr 2023, 01:02
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One possible way to deal with S1:

√36x= integer

Squaring both sides:

36x = integer^2 (perfect square)

In other words, 36x is also a perfect square, which means all of its primes have even exponents

We know that 36 = 2^2 x 3^2

Therefore, x must also contain primes in even powers, thus making it a perfect square and its square root an integer
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If x is a positive integer, x^(1/2) an integer? 17 Jan 2025, 10:50
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