Latest CTS Aptitude Question SOLUTION: 2^74+2^2058+2^2n,what is the value of n for which the expression is a perfect square?

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2^74+2^2058+2^2n,what is the value of n for which the expression is a perfect square?

Asked In CTS krishnaraj.v (9 years ago)

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PRADEEP KUMAR  (8 years ago)

(a^2)+2ab+(b^2)
(2^37)^2+(2*2^37*2^n)+(2^2n)=(2^74)+(2*2^37*2^2020)+(2^4040)
hence by comparision
2n=4040 ; n=2020

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Sakshi Gupta  (8 years ago)

23 is the correct ans
4(1066+n)
n=23
4(1066+23)
4(1089)
which is a perfect square of 33*2=66

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Saragadam Beerendra Babu  (8 years ago)

a^2 +2ab+b^2

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satish kumar  (4 years ago)

2^74 + 2^2058 + 2^2n =(2^37)^2 + (2^1029)^2 + (2^n)^2
now, if we put 2^37 = a ; 2^1029 = b; Then for the above expression to be perfect square 2^2n must be equal to (2*a*b)= 2*(2^37)*(2^1029);
==> 2^2n = 2^(1067)
==> 2n = 1067 ,
but this case is not possible since R.H.S is an odd integer whereas L.H.S is an even integer.
So , the above mentioned case can't hold.

Now,if we put 2^37 = a; 2^n = b ; So, for the given expression to be perfect square 2^2058 = (2*a*b)= 2*(2^37)*(2^n) = 2^(n+38);
So, 2058 = (n+38)
=> n = 2020
So,The answer is n = 2020

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