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If x is a positive integer, is x^1/2 an integer [#permalink]
 10 Jan 2010, 00:55
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If x is a positive integer, is \sqrt{x} an integer? (1) \sqrt{4x} is an integer. (2) \sqrt{3x} is not an integer. the explanation says since sqrt (4x) is an integer ,it follows that 4x must be square of an integer and so x must be
square of an integer and therefore sqrt (x) is an integer .
i was trying to solve it this way sqrt(4x) =integer ,=> 2. sqrt (x) =integer => sqrt (x) =integer/2 =integer or non integer for example if 2 .sqrt (x) = 4 ,=> sqrt (x) =2 and so sqrt (x) is integer
but if 2. sqrt (x) =3 ,=> sqrt (x) =3/2 and so sqrt (x) is non integer
friends please help me in pointing out whee i am going wrong.
Thanks in advance ..
Last edited by Bunuel on 16 Dec 2012, 08:31, edited 1 time in total.
Renamed the topic, edited the question and added OA.
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]
09 Jan 2011, 09:52
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fluke wrote: I didn't get the explanation:
if root(4x) is an integer fact.
Then 2 * root(x) is an integer. So what!!!!!!!!!!
This doesn't mean that root(x) should be an integer. Because root(x) can be 1.5 and yet won't distort the fact that 2 * 1.5 = 3 is an integer.
So, if x = 2.25(a non integer) root(x)=1.5 and 2*1.5=3 is an integer.
if x=4(an integer) root(x)=2 and 2*2=4 is also an integer.
So, statement one would be true for two values of x (2.25 and 4).
root(2.25) is 1.5, not an integer. root(4) is 2, an integer. This statement is insufficient to conclude whether root(x) is an integer.
What's wrong with my explanation?? You forgot that x is a positive integer, so \sqrt{x} cannot equal to \frac{integer}{2}. Generally \sqrt{integer} is either an integer or an irrational number. Complete solution: If x is a positive integer, is sqrt(x) an integer(1) sqrt(4x) is an integer . (2) sqrt(3x) is not an integer . If x=integer, is \sqrt{x}=integer? (1) \sqrt{4x} is an integer --> 2\sqrt{x}=integer --> 2\sqrt{x} to be an integer \sqrt{x} must be an integer or integer/2, but as x is an integer, then \sqrt{x} can not be integer/2, hence \sqrt{x} is an integer. Sufficient. (2) \sqrt{3x} is not an integer --> if x=9, condition \sqrt{3x}=\sqrt{27} is not an integer satisfied and \sqrt{x}=3 IS an integer, BUT if x=2, condition \sqrt{3x}=\sqrt{6} is not an integer satisfied and \sqrt{x}=\sqrt{2} IS NOT an integer. Two different answers. Not sufficient. Answer: A. Hope it's clear.
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]
16 Dec 2012, 08:37
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]
10 Jan 2010, 08:42
if x is a positive integer ,is sqrt (x) an integer
(1) sqrt(4x) is an integer .
(2) sqrt (3x) is not an integer .
the explanation says since sqrt (4x) is an integer ,it follows that 4x must be square of an integer and so x must be
square of an integer and therefore sqrt (x) is an integer .
i was trying to solve it this way sqrt(4x) =integer ,=> 2. sqrt (x) =integer => sqrt (x) =integer/2 =integer or non integer for example if 2 .sqrt (x) = 4 ,=> sqrt (x) =2 and so sqrt (x) is integer
but if 2. sqrt (x) =3 ,=> sqrt (x) =3/2 and so sqrt (x) is non integer
yes.
sqrt(x) is an integer.
how i worked it out :
sqrt(4x) is an integer ==> 2 * sqrt(x) is an integer
in order that the product of 2 and sqrt(x) be an integer, sqrt(x) must either be
1) an integer
2) exactly half of an integer. i.e a number like 0.5, 1.5, 2.5 etc etc
you worked that out as well.
sqrt(x) = integer/2
we also know x is an integer. is there any integer whose square root is a half of an integer ?
no!
therefore the only other alternative is sqrt(x) is a whole integer.
hope that helped
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]
11 Jan 2010, 06:50
Hi Janani ,
Thanks for the explanation ,do we have a rule like we can't have a sqrt (x)=integer /2,
or its by observation ...
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]
09 Jan 2011, 09:40
I didn't get the explanation: if root(4x) is an integer fact. Then 2 * root(x) is an integer. So what!!!!!!!!!! This doesn't mean that root(x) should be an integer. Because root(x) can be 1.5 and yet won't distort the fact that 2 * 1.5 = 3 is an integer. So, if x = 2.25(a non integer) root(x)=1.5 and 2*1.5=3 is an integer. if x=4(an integer) root(x)=2 and 2*2=4 is also an integer. So, statement one would be true for two values of x (2.25 and 4). root(2.25) is 1.5, not an integer. root(4) is 2, an integer. This statement is insufficient to conclude whether root(x) is an integer. What's wrong with my explanation??
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]
10 Jan 2011, 01:32
good explanation Bunuel and you are right in saying that I carelessly overlooked the fact that x was a positive integer... thanks ~fluke
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]
19 Jul 2011, 06:29
Quote: then sqrt(x) can not be integer/2 I think sqrt(x) can be integer/2 as long as (integer/2) itself is an integer  i.e. that integer is multiple of 2. I think that is what bunuel meant. And the answer remains same.
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]
19 Jul 2011, 08:18
a (4x)^1/2 = 2 * (x)^1/2 = integer thus sufficient. b (3x)^1/2 = 3 * (x)^1/2 where x can be 4 or 5. hence not sufficient. thus A it is.
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Re: if x is a positive integer ,is sqrt (x) an integer (1) [#permalink]
15 Dec 2011, 17:59
Hi,
"You forgot that x is a positive integer, so sqrt(x) can not equal to integer/2 . Generally sqrt(x) is either an integer or an irrational number."
Am i missing something here?
sqrt(4) = 4/2 = 2. This is integer by 2 right? (Doesnt matter much in this problem's case)
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Re: if x is a positive integer ,is sqrt (x) an integer (1) [#permalink]
03 Jan 2012, 13:06
Oh nice problem. Took quite some time to answer it but got A. Explanation by Bunuel is more than sufficient to understand the solution.
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Re: if x is a positive integer ,is sqrt (x) an integer (1) [#permalink]
03 Jan 2012, 14:31
Both equations and number plugging helps here. 1. sqrt(4x) = integer, that is 2 x sqrt(x) = integer. For this equation to be true, sqrt(x) has to be an integer. SUFFICIENT. If number plugging, use 4x4 and 4x9 combinations 2. sqrt(3x) = frac. Here, sqrt(3) x sqrt(x) = frac, that is frac x sqrt(x) = frac. Difficult to determine if this relationship can infer sqrt(x) as integer. So, let's go with number plugging. sqrt(3x4) = 12 satisfies, and sqrt(4) = integer. sqrt(3x5) = 15 satisfies, but sqrt(5) = frac. So, insufficient. +1 for A.
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]
15 Dec 2012, 11:21
You forgot that x is a positive integer, so \sqrt{x} can not equal to \frac{integer}{2}. Generally \sqrt{integer} is either an integer or an irrational number.
Complete solution:
If x is a positive integer, is sqrt(x) an integer
(1) sqrt(4x) is an integer .
(2) sqrt(3x) is not an integer .
If x=integer, is \sqrt{x}=integer?
(1) \sqrt{4x} is an integer --> 2\sqrt{x}=integer --> 2\sqrt{x} to be an integer \sqrt{x} must be an integer or integer/2, but as x is an integer, then \sqrt{x} can not be integer/2, hence \sqrt{x} is an integer. Sufficient.
(2)\sqrt{3x} is not an integer --> if x=9, condition \sqrt{3x}=\sqrt{27} is not an integer satisfied and \sqrt{x}=3 IS an integer, BUT if x=2, condition \sqrt{3x}=\sqrt{6} is not an integer satisfied and \sqrt{x}=\sqrt{2} IS NOT an integer. Two different answers. Not sufficient.
Answer: A.
Hope it's clear.[/quote]
Bunuel - can you further explain your explanation for Statement 1? I am confused because if x is a positive integer, sqrt(x) can equal an integer/2 if for example x was equal to 4. 4 is a positive integer and the square root of 4 is equal to 4/2. I think I am missing the overall takeaway, can you help clarify? Thanks!
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Re: if x is a positive integer ,is sqrt x an integer
[#permalink]
15 Dec 2012, 11:21
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