self
Maths Puzzle
for which of the following n is the number 2^74 + 2^2058 + 2^2n is a perfect square
Read Solution (Total 6)
-
- 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 - 12 years agoHelpfull: Yes(30) No(6)
- 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 - 12 years agoHelpfull: Yes(21) No(8)
- 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 - 6 years agoHelpfull: Yes(2) No(1)
- 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 - 8 years agoHelpfull: Yes(0) No(3)
- 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 - 7 years agoHelpfull: Yes(0) No(2)
- 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 - 7 years agoHelpfull: Yes(0) No(2)
self Other Question
In a sequence of integers, A(n)=A(n-1)-A(n-2), where A(n) is the nth term in the sequence. n is an integer and n>=3, A1=-1,A2=1. Calculate S(1000) where S(1000) is the sum of the first 1000 terms
1,3,7,28,33,__,__