Is the square root of x is an integer?

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

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

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.

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!

Is there any way for me to know the level of the question? I am really curious when I practice GMAT

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

The units digit of x is 2.

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