Latest TCS Aptitude Question SOLUTION: 2^1096+2^2248+2^2n for what value of n it become perfect square a)2008 b)2010 c)2011 d)2012

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2^1096+2^2248+2^2n
for what value of n it become perfect square
a)2008
b)2010
c)2011
d)2012

Asked In TCS madhavaram mahendra (11 years ago)

Unsolved Read Solution (4)

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semanti dey  (11 years ago)

answer will be 2011.
see unit digit of 2^1096 is 6((2^4)^274)
and unit digit of 2^2248 is 6 also...
so sum of those digit is 12..which is not a prfct sqr..
now if unit digit of 2^2n becomes 4..den 12+4=16 can be a prfct sqr..
if we take the option 2008.2^2n will be 2^4016..unit digit of this is 2..
which is not correct..if we take 2011..2^2n will be 4^2011..unit digit of ds is 4.
now the sum is 12+4=16..which is a prfct sqr..

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Rahul  (11 years ago)

wrong question
(a+b)^2=a^2 + b^2 + 2ab
if its around 2000 then
it cant be 2ab(2^(548+1124+1))
also 2^1096 cant be 2ab(

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Rahul  (11 years ago)

1096

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pumang jha  (11 years ago)

1096=8*137
2248=8*281
so,
2008=8*251....

therefore,
(A) is the ans

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