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by EMAN » Mon Sep 14, 2009 10:00 am
If x^2 is divisible by 216, what is the smallest possible value for positive integer x?
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by xcusemeplz2009 » Mon Sep 14, 2009 10:36 am
36

factors of 216=2^2*3^2*2*3

x^2=(2*3*2*3)^2 is the smallest no. div by 216
x=2*3*2*=36
It does not matter how many times you get knocked down , but how many times you get up
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Correct

by EMAN » Mon Sep 14, 2009 4:45 pm
Correct. Could you explain how that works? I was having trouble understanding the Manhattan guide.
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by xcusemeplz2009 » Mon Sep 14, 2009 5:58 pm
its basically to understand that x^2 in oredr to be perfactly divisible by 216 it has to satisfy two condition
x^2=216*n( where n=1,2,3,4,.....)
secondly 216*n has t be a perfact square no., further factorising it 216*n= 36*6*n; if n=6 then this number can be perfact square no;putting n=6
x^2=36*6*6 or x=sqrt(36*36)=36
It does not matter how many times you get knocked down , but how many times you get up
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