Is x a multiple of 12?

Only an integer can be odd. i.e. if sqrt{x-3} is odd then it must be a number. So 18 do not satisfies.

hmm.. decimals cannot be even or odd.. thanks!

Statement 1:

Let sqrt(x-3) =n with n being odd.
<=> x-3 = n^2 <=> x = n^2 + 3

Going through odd integers we get x = 4, 12, 28, 52, 84, 124, 172 etc.

=> INSUFF.

Statement 2: x is a multiple of 3 => INSUFF.

Both statements combined:

Looking at the list of possible values for x above, we see that all values that are divisible by 3 are divisible by 4, too. Therefore, these values are also a multiple of 12.

=> Answer should be C.

Hi,
the OA is given as "d" in this question however I think it should be "c" as others have also pointed out?

_____________________
Edited the OA. Thank you.

Is x a multiple of 12?

(1) \(\sqrt[]{x−3}\) is odd
=> \(\sqrt[]{x−3}\) = a => a= odd integer
=> \(x-3 = a^2\)
=> \(x= a^2+3\)

=> here x is an integer
For a as an odd integer, for a=1,3,5,7,9.... value of x = 4,12,28,52,84.......
Insufficient

(2) x is a multiple of 3
=> x is an integer and x =3b => here b is an integer => x=3,6,9,12,15,18,21,24......
Clearly Insufficient

1+2
\(x= a^2+3\) and x =3b , here a is an odd integer and b is an integer
=> \(3b=a^2+3\)
=> b= \(\frac{a^2}{3}\)+1
Here as 'b' need to be an integer 'a' should be multiple of 3

=> a is odd and multiple pf 3
=> \(x= a^2+3\) , a=3,9,15 => x=12,84,228......
=> x is multiple of 12

Answer: C

Both taken together

As x is a multiple of 3, we need to see whether x is a also a multiple of 4.
Let's say \(\sqrt{x-3} = k\)
\(x = k^2 + 3 = (k - 1)(k+1) + 4\)

Because k is an odd number(we have that condition from \(\sqrt{x-3}\) is an odd number), that means \((k - 1)(k+1) + 4\) is multiple of 4. So we have x is multiple of 3 and 4, so x is multiple of 12

Official Explanation:
Since statement (1) indicates that \(\sqrt{x-3}\) is odd and the square root sign implies a positive answer, list 1, 3, 5, 7, 9, etc.
Notice that you're picking values for \(\sqrt{x-3}\) , not x. It would be far too much work to test different values for x to determine which make \(\sqrt{x-3}\) odd, and you could potentially miss some values that fit the statement. Do not plug in numbers for x here! Instead, list consecutive odd values for \(\sqrt{x-3}\), a quick and easy process. Then solve for x in each case.
For this problem, your work on paper may look something like this:

(1) INSUFFICIENT: \(\sqrt{x-3}\) = odds = 1, 3, 5, 7, 9, etc.
x − 3 = 1, 9, 25, 49, 81, etc.
x = 4, 12, 28, 52, 84, etc.
Is x divisible by 12? Maybe. For example, 12 is, while 28 is not.

(2) INSUFFICIENT: x = multiples of 3 = 3, 6, 9, 12, 15, etc.
Is x divisible by 12? Maybe. For example, 12 is, while 15 is not.

(1) AND (2) SUFFICIENT: Combine these statements by selecting only the values for x that are in both lists. On your paper, circle the following values: x = 12 and x = 84. These are the values calculated in statement (1) that fit the criteria in statement (2). This seems to be SUFFICIENT—the values for x that fit both statements are multiples of 12. At this point, if you wanted to check another value, you could, or you could go with the trend, which is almost always going to be right after testing this many cases.
The correct answer is (C).

Is the red part ok Bunuel?
Thanks__

Decimal value numbers are neither even nor odd. So saying that \sqrt{x-3} is odd implicitly means it is an interger.