Latest TCS Aptitude Question SOLUTION: 2^74 + 2^2058 + 2^2n. what is the value of n make it perfect square?

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2^74 + 2^2058 + 2^2n. what is the value of n make it perfect square?

Asked In TCS sowjanya (9 years ago)

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RAKESH  (9 years ago)

(a+b)^2 = a^2 + 2ab + b^2 :: 2^74 + 2^2058 + 2^2n
compare it
a^2 = 2^74 => a = 2^37
& b^2 = b^2n => b =2^n
2ab = 2^2058 => 2 * 2^37 * 2^n
=> 2^2058 = 2^(38+n)
=> 38+n = 2058
=> n = 2020

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varun peddireddy  (9 years ago)

2^74 + 2^2058 + 2^2n = ((2^37)^2) +(2* (2^37) * (2^2020))+((2^n)^2)
this is of the form (a+b)^2= a^2+2ab+b^2
where a=(2^37) and b= (2^2020)=(2^n)
therefore it can be written as ((2^37)+(2^2020))^2 which is a perfect square
n=2020

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ranjan kumar  (9 years ago)

so,looking at the first glance we see that it is something relative to (a+b)^2

hmm.. now (a+b)^2 = a^2 + b^2 + 2ab
now match it with the question..
2^74 = a^2 so what should be a ?? 2^37
similarly b = 2^n
now 2ab = 2* 2^37 * 2^n(putting the respective values)
and this 2ab = 2^2058
so adding the powers you get 1+37+n
and that will be equal to 2058.
ans nikal lo yaar..

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Venkidesh Sreekumar  (9 years ago)

please give the options because the other solutions are right abut if you take a^2=2^74 then a = 2^37 and if b^2=2^2058 then b = 2^1029, then 2ab = 2^1067. therefore 2n =1067 and n can be 533.5. so the answer can be either 2020 or 533.5

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