A Radical Radical

[m.malini]

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)

I'm looking for a way to approach and solve this problem. Thanks in advance!

[goal.ambitions]

My approach
we need to find sq. root of 288xk
Since x is divisible by 6, let's assume that x=6m
so 288xk=288*6mk=(2^6)*(3^3)*mk
If we take square root of above then 24 will come out & there will be 3mk under the square root.

Now looking at the options one by one
A) 24k*sqrt(3)
so 3mk=3, so mk=3, it means m=1,k=1, possible

B) 24*sqrt(k)
so 3mk=k means 3m=1...NOT POSSIBLE AS m is an integer

C) 24*sqrt(3k)
3mk=3k..posssible

D) 24*sqrt(6k)
3mk=6k..possible

E) 72*sqrt(k)
72*sqrt(k)=24*sqrt(9k)
means 3mk=9k.possible

The answer would be B
PLEASE ADVISE OA

[jnelson0612]

VERY nice goal.ambitions. This is an excellent explanation.

Thank you!