Latest TCS Aptitude Question SOLUTION: (2^1096 + 2^2248 + 2^2n).find the value of n so that the value will be perfect square (a)2011(b)2012(c)2010(d)2008

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(2^1096 + 2^2248 + 2^2n).find the value of n so that the value will be perfect
square (a)2011(b)2012(c)2010(d)2008

Asked In TCS viru (11 years ago)

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deviprasad gullipalli  (11 years ago)

to be a perfect square we can rite given term into foll. format (a+b)^2=a^2+2*a*b+b^2
(2^548)^2 + 2*(2^548)*(2^n)+(2^n)^2
comparing it ith given term,,,, 2248=1+548+n
n=1699 is correct

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viru  (11 years ago)

(2^1096 + 2^2248 + 2^2n)
kanak saxena ji there is + between numbers not * how you give your answer

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kanak saxena  (11 years ago)

ans is (a) by hit & trial method.
to be a perfect sq. a nos power should be multiple of 2, obtion b,c,d satisfy this statement.
2^(1096 + 2248 + 2n) = 2^(3344+ 2n)....(1)
n= 2011, 2012 , 2010 , 2008.
by putting d value of n in eqt 1, we get 7366 , 7368 ,7364 and 7360 resp..
by dividing this sum by d cyclicity of 2 i.e 4 we get integer 4m obtion b, c,and d.
oddmann out, obtion 'a' is differ 4m other , so answer is a. ;-)

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