Is x a multiple of 12?

Question

(1)  \sqrt{x-3}  is odd
(2) x is a multiple of 3

Nikkb · Accepted Answer

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

Sudhanshuacharya · Answer

I believe the answer should be C,
S1 tells us that X is an even number greater than 3 so we have 4,12,28. (As \sqrt{(x-3)} is also an odd number So not sufficient
S2 tells us that X is a multiple of 3. so we have 3,6,9... so not sufficient.

S1 and S2 tells us that X is an even multiple of 3 that is 12,36,

So IMO C.

What is the OA

dreambeliever · Answer

OA is C.

However, S1 says \sqrt{x-3} is odd. It does not however say that it is an integer or not.

for eg. the number 18 will satisfy both the both S1 and S2. As Sq root of 15 is odd and 18 is a multiple of 3. In fact sq of any odd number will be odd. Unless we assume that the statement "sq root of a number is odd" means that the number is a perfect sq. 
Can someone please clarify if that this is a valid assumption that we can make for all gmat questions?

maliyeci · Answer

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.

dreambeliever · Answer

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

dmnk · Answer

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.

Alabama · Answer

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

TheUltimateWinner · Answer

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) .

liammoyn · Answer

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

Roopakrai · Answer

X-3 = (odd)^2
X= odd^2+3
X= Even

X= multiple of 3

Combining the both, X = even multiple of 3 
X=6,12,18,24,30

Shouldn’t it be E ?

Posted from my mobile device

Bunuel · Answer

6, 18, 24, 30, and so on do not satisfy odd^2 + 3.

odd^2 + 3 gives: 4,  12 , 28, 52,  84 , ...

Multiples of 3 among these are also multiples of 12.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Is x a multiple of 12?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

Is x a multiple of 12?

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards