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 (11 years ago)

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Akanksha Narula  (11 years ago)

find the unit digit of 2^1096
wch cms out 2 b = 6
nw again unit digit of 2^2248= 6
add these unit digits= 6+6=12
now we knw any no. wch ends wd 2 is not a perfect square.

consider last term: 2^2n= 4^n
if n=2011 then unit digit of 4^2011= 4
add 12+4 = unit digit= 6
it can make a perfect square

if n=2012, then unit digit of4^2012 = 6
adding 12+6+unit digit= 8
no.ending wd 8 can not b a perfect square.
||ly wd the other options. :)
so ans is 2011

Is this solution Helpfull? Yes (18) | No (7)  

vivek chauhan  (11 years ago)

ansr:2011
just calculate unit digit in b,c,d it will come 3 nd the number which end with 3 cant be perfect square

Is this solution Helpfull? Yes (13) | No (36)  

onkar nagarkar  (11 years ago)

to find the value of n, just look at options first
2011, 2012,2010 are not divisible by 8 but 2008 is divisible by 8......
also 1096 and 2248 is also divisible by 8
hence ans is (Id) 2008

Is this solution Helpfull? Yes (3) | No (13)  

ABHINAV SINGH  (11 years ago)

ans is 2012
as 2^1096 give unit digit 6
2^2248 give umit digit 6
and now we calculate unit digit of third i.e 2^2*(2011) give unit digt 2
so (6*6*2= 72 here unit digit of 2 is not a perfect sq)
so take now 2^2*2012 which give unit digit 4 so now (6*6*4=144)
hence 144 unit digit is 4 so perfect sq..

Is this solution Helpfull? Yes (1) | No (9)  

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