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Divisibility Question

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by scoobydooby » Mon May 11, 2009 10:36 pm
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
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Re: Divisibility Question

by Pranay » Tue May 12, 2009 1:56 am
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
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Re: Divisibility Question

by scoobydooby » Tue May 12, 2009 2:01 am
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
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Re: Divisibility Question

by Pranay » Tue May 12, 2009 2:14 am
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.. :)
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