Problem Solving

X is divisible by 144. 3 sqrt x is an integer, then which of the following 3 sqrt x definitely divisible by? (Choose all that apply)
A) 4
B) 8
C)9
D)12

[spoiler]A & D[/spoiler]

Thanks!

i get A, C and D

x/144=k (integer) 144=2^4*3^2
the minimum that x can be 2^4*3^2 (x not zero, as sq. root defined over positive numbers)

if minumum x=2^4*3^2; sq root x is 2^2 *3

3 sqrtx =2^2*3^2 at aleast.

A. 2^2*3^2 divisible by 4
B. 2^2*3^2 not divisible by 8
C. 2^2*3^2 divisible by 9
D. 2^2*3^2 divisble by 12

so A, C and D

I get all the options..

My working as under,

x is divisible by 144 => x is a multiple of 144, may be 144*n

3sqrt(x) = 3*sqrt(144*n) = 36sqrt(n).

Since, 36 is divisible by all the options given, I get all the options as answer.

Please correct my working.

Pranay

Pranay wrote:I get all the options..

My working as under,

x is divisible by 144 => x is a multiple of 144, may be 144*n

3sqrt(x) = 3*sqrt(144*n) = 36sqrt(n).

Since, 36 is divisible by all the options given, I get all the options as answer.

Please correct my working.

Pranay

except that 36 is not divisible by 8

scoobydooby wrote:

Pranay wrote:I get all the options..

My working as under,

x is divisible by 144 => x is a multiple of 144, may be 144*n

3sqrt(x) = 3*sqrt(144*n) = 36sqrt(n).

Since, 36 is divisible by all the options given, I get all the options as answer.

Please correct my working.

Pranay

except that 36 is not divisible by 8

Thanks Scoobydoo for correcting.

That is the silliest mistake that one can make..