If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of sqrt (288kx) ?

A) 24k sqrt(3)
B) 24 sqrt(k)
C) 24 sqrt(3k)
D) 24 sqrt(6k)
E) 72 sqrt(k)

What would be your best approach to solve this problem quickly and accurately ?
Thanks.

Please feel free to state any secondary strategies you would use to answer this incase you were running out of time, or came stuck on this problem.

Thanks in advance.

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of sqrt (288kx) ?

A) 24k sqrt(3)
B) 24 sqrt(k)
C) 24 sqrt(3k)
D) 24 sqrt(6k)
E) 72 sqrt(k)

Take X = 6N where N=1,2,3 ...... and solve for different values of SQRT(288KX) = SQRT (2 X 3 X 2 X 3 X 3 X4 X4 KN) = 24 SQRT ( 3 X KN)
and put K & N=1,2,3

Please make sure to follow protocol: your subject should be the first 5 to 8 letters of the problem. Thanks.

nag broke 288 down into its prime factors - and, yes, you should do this to discover what you can pull out from the SQRT sign.

You can do this one of two ways: use a factor tree and divide by whatever small numbers strike you, or notice 24 and 72 in the answers. 24 is in four choices, so chances are you can pull out a 24. So you could also divide 288 by 24 to shortcut the factor tree - just depending on what method you like better.

288/24 = 12. So one factor pair is 24*12 which breaks down to 2*2*2*3*2*2*3. I have five 2's, so I can take out four, and I have two 3's, so I can take out both. That gives me 12*SQRT2. Then, of course, I was told that x is divisible by 6, so I can use nag's method: x=6n, where n is a positive integer. Now I've got another 2 and 3 under the square root sign to go with the 2 already there. Pull out another pair of 2's to get 24*SQRT3. And, of course, we've got our variables under the SQRT sign: k and n. So now we're down to:
24*SQRT(3kn)

So, let's see, which of the answer choices COULD I create?
A) If n = k, then I can pull k out of the SQRT, which gives me 24kSQRT3. Eliminate A.
B) Hmm. In this one, I only have 24 out front, so nothing else gets pulled out. I start with 3kn under the square root sign and this choice shows only k in there. How could I get rid of the 3 and the n? Well, they could cancel each other out. For that, n would have to equal 1/3. But n has to be an integer... so I can't make this one. B is the right answer.
C) If n = 1, then this is what I get
D) If n = 2, then this is what I get
E) If n = 3, then I can pull another 3 out of the SQRT and I've got 24*3*SQRTk or 72SQRTk. Eliminate E.

If you're really short on time and just have to guess:
- don't guess E. It's too obviously "different" than the rest (72 instead of 24). They're asking what "cannot" be true and too many people who don't know would guess the obviously odd one out.
- I'd eliminate A for the same reason as above (moving k outside the square root sign)
After that, I'd just guess.

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

thanks Stacey. as you said, substituting 6N for X is a major step in solving this problem and which is where i was stuck terribly. as for the educated guessing technique, i think this problem is kinda gamble. i would either solve or guess randomly.

also, what is the protocol you mentioned earlier? did you mean to say the 1st 5 words as in "If k and x are..?"

_________________
Ashish
Share why wrong answer(s) are not right than just stating why the right one is.

Correct, the first five words.

_________________
Jamie Nelson
ManhattanGMAT Instructor