Latest self Aptitude Question SOLUTION: for which of the following n is the number 2^74 + 2^2058 + 2^2n is a perfect square

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for which of the following n is the number 2^74 + 2^2058 + 2^2n is a perfect square

Hardik Patel (11 years ago)

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SDeep Bhattacharya  (11 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

Is this solution Helpfull? Yes (30) | No (6)  

Jamil Ahmed  (11 years ago)

We know that a^2 + 2ab + b^2 = (a+b)^2
Here, 2^74 + 2^2058 + 2^2n = (2^37)^2 + 2*2^37*2^2020 + (2^n)^2
If the number is a perfect square, then 2^n should equal 2^2020.
Hence, n=2020

Is this solution Helpfull? Yes (21) | No (8)  

Swapnil Sen  (5 years ago)

Given expression 2^74 + 2^2058 + 2^2n

We can write the above expression in the form of (a+b)^2

2^74 + 2^2058 + 2^2n

=(2^37)^2 + 2 X 2^37 X 2^n + (2^n)^2

=(2^37)^2 + 2^(38 + n) + (2^n)^2

So , we can write

(38 + n) = 2058

=> n = 2020

Is this solution Helpfull? Yes (2) | No (1)  

ankit das  (8 years ago)

2^74 +2^2058+2^2n = K^2
2^74 +2^2058+2^2n = (2^37)^2+2^2058+(2^n)^2
We try to write this expression as (a+b)^2=a^2+2ab+b^2
Now a = 2^37, 2ab = 2^2058 and b = 2^n
Substituting the value of a in 2ab, we get b = 2020

Is this solution Helpfull? Yes | No (3)  

Souvik Saha  (6 years ago)

2^74 +2^2058+2^2n =
K
2
K2
2^74 +2^2058+2^2n =
(
2
37
)
2
+
2
2058
+
(
2
n
)
2
(237)2+22058+(2n)2
We try to write this expression as
(
a
+
b
)
2
=
a
2
+
2
a
b
+
b
2
(a+b)2=a2+2ab+b2
Now a =
2
37
237, 2ab =
2
2058
22058 and b =
2
n
2n
Substituting the value of a in 2ab, we get b = 2020

Is this solution Helpfull? Yes | No (2)  

Souvik Saha  (6 years ago)

2^74 +2^2058+2^2n =
K
2
K2
2^74 +2^2058+2^2n =
(
2
37
)
2
+
2
2058
+
(
2
n
)
2
(237)2+22058+(2n)2
We try to write this expression as
(
a
+
b
)
2
=
a
2
+
2
a
b
+
b
2
(a+b)2=a2+2ab+b2
Now a =
2
37
237, 2ab =
2
2058
22058 and b =
2
n
2n
Substituting the value of a in 2ab, we get b = 2020

Is this solution Helpfull? Yes | No (2)  

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