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# Is the square root of x is an integer?

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Senior Manager
Joined: 06 Aug 2011
Posts: 341
Is the square root of x is an integer?  [#permalink]

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Updated on: 08 Mar 2014, 07:22
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Difficulty:

35% (medium)

Question Stats:

67% (01:24) correct 33% (01:19) wrong based on 293 sessions

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Is the square root of x is an integer?

(1) The unit digit of x is 2
(2) x is divisible by 3.

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Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Originally posted by sanjoo on 07 Mar 2014, 05:54.
Last edited by Bunuel on 08 Mar 2014, 07:22, edited 1 time in total.
Renamed the topic and edited the question.
Manager
Joined: 01 May 2013
Posts: 61
Re: is sqroot x is an interger?  [#permalink]

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07 Mar 2014, 08:41
1
Only the units digit determines the units digit of the product. Check 10 cases.

(1) The units digit of x is 2.

Units digit --> Units digit of the square

1 --> 1
2 --> 4
3 --> 9
4 --> 6
5 --> 5
6 --> 6
7 --> 9
8 --> 4
0 --> 0

The units digit of a perfect square can never be 2. Statement 1 is sufficient to answer the question.

(2) x is divisible by 3

x = 3 No
x = 9 Yes

Statement 2 is NOT sufficient. The answer is A.
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Re: Is the square root of x is an integer?  [#permalink]

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09 Mar 2014, 02:35
1
sanjoo wrote:
Is the square root of x is an integer?

(1) The unit digit of x is 2
(2) x is divisible by 3.

Sol: We need to find whether $$\sqrt{x}$$ or x=I^2 where I is an Integer

St 1: Unit digit of x is 2...
Consider x=2^(Some positive integer ) so we get x=2,4,8,16 and 32 (2^5)

$$\sqrt{32}$$ is not an integer
Now x=3^ (some positive number, it will not end in 2 in any case)....
Consider x=4 ^(Some positive number) so we get x=4,16,64,256 (Ends in 4 or 6 )
Consider x=6 ^(Some positive number)...Always ends in 6
Consider x=8^(Some power) we get x=8,64,512,---6,----8
Now x=512 then $$\sqrt{512}$$ is not an integer

Based on above we can say that$$\sqrt{x}$$ is not an integer.We did not consider value of x=5,7,9 (^ some positive number) because it can not end in zero.

St 1 is sufficient

St2: Possible value X=3,6,9,12....SO if X=9 then$$\sqrt{X}$$ is an Integer but if x=6 then not.

Ans is A
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Re: Is the square root of x is an integer?  [#permalink]

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10 Mar 2014, 23:47
sanjoo wrote:
Is the square root of x is an integer?

(1) The unit digit of x is 2
(2) x is divisible by 3.

This question is quite straight forward.
Remember that unit's digits of perfect squares can never be 2/3/7/8.
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Re: Is the square root of x is an integer?  [#permalink]

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11 Mar 2014, 00:09
1
Ans A:

as we can see

square of any number will not end in 2
hence A is straightaway an answer that root of x can never be an int
Senior Manager
Joined: 08 Apr 2012
Posts: 375
Re: Is the square root of x is an integer?  [#permalink]

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29 Jun 2014, 04:33
1
VeritasPrepKarishma wrote:
sanjoo wrote:
Is the square root of x is an integer?

(1) The unit digit of x is 2
(2) x is divisible by 3.

This question is quite straight forward.
Remember that unit's digits of perfect squares can never be 2/3/7/8.

Hi Karishma,

Do you have an algebraic approach to prove this?
I don't think I'll remember it on test day and was wondering if I can derive this on my own...?
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Re: Is the square root of x is an integer?  [#permalink]

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29 Jun 2014, 21:39
1
ronr34 wrote:
VeritasPrepKarishma wrote:
sanjoo wrote:
Is the square root of x is an integer?

(1) The unit digit of x is 2
(2) x is divisible by 3.

This question is quite straight forward.
Remember that unit's digits of perfect squares can never be 2/3/7/8.

Hi Karishma,

Do you have an algebraic approach to prove this?
I don't think I'll remember it on test day and was wondering if I can derive this on my own...?

The units digit of any integer can take any one of the 10 distinct values - 0 to 9.

When you square an integer, the last digit of the integer multiplies with itself to give the last digit of the square.
Check out that last digit when you square these 10 distinct digits:

0^2 = 0
1^2 = 1
2^2 = 4
3^3 = 9
4^4 = 6
5^5 = 5
6^6 = 6
7^2 = 9
8^2 = 4
9^2 = 1

So you will never get 2, 3, 7 or 8 when you square an integer.
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Karishma
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Re: Is the square root of x is an integer?  [#permalink]

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16 May 2017, 16:30
aman89 wrote:
Ans A:

as we can see

square of any number will not end in 2
hence A is straightaway an answer that root of x can never be an int

I also come up with the option A, but then my brain gets confused and pick C instead. This is also an experience for new members, cheer!
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Posts: 10
Location: United States (CA)
Concentration: Real Estate, Nonprofit
GPA: 3.85
Re: Is the square root of x is an integer?  [#permalink]

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16 May 2017, 16:31
Is there any way for me to know the level of the question? I am really curious when I practice GMAT
Intern
Joined: 15 May 2017
Posts: 3
Re: Is the square root of x is an integer?  [#permalink]

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16 May 2017, 23:15
Is the square root of x is an integer?
(1) The unit digit of x is 2
(2) x is divisible by 3.

Ans:
Number with unit place ranging from 0-9 will never give 2 at the unit place if multiplied by itself , ( 1*1=1, 2*2=4, ..... , 9*9=81)
Hence (1) statement tells that square root of x is not an integer.

Now, statement (2) would take just few hit and trials, if x is the multiple of 3 then it could be 3, 6, 9. And square roots of 3 & 6 are not integers.
Hence (2) statement is insufficient.

Good Day
Re: Is the square root of x is an integer? &nbs [#permalink] 16 May 2017, 23:15
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