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(#M40012524) TCS Numerical Ability question Keep an EYE

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

Asked In TCS RAHU SHARMA (12 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

thrishna  (12 years ago)

(2^37)2+2(2^37)(2^n)+(2^n)2= 2^74+2^2058+2^2n
2^(38+n)=2^2058
38+n=2058
n=2058-38
n=2020
ANS:2020

Is this solution Helpfull? Yes (31) | No (7)  

vishnu priya vellanki  (12 years ago)

as (a+b)^2 is perfect square,
apply (a+b)^2=a^2+b^2+2ab formula,
compare given equation with the above formula,
2^74 can be written as, (2^37)2
2^2n can be written as, (2^n)2
so,a=2^37,b=2^n
2^2058=2ab=>2*(2^37)*(2^n)
now,compare powers on both sides, then u'll get 2058=1+37+n
so,n=2020
2058=

Is this solution Helpfull? Yes (10) | No   

lala  (12 years ago)

niki
if 2020 is not in answer
leave the ques this is a dummy question

Is this solution Helpfull? Yes (2) | No (3)  

sasi  (12 years ago)

(2^37)^2+2*(2^37)*(2^n)+((2^n)^2)....a square+2*a*b+b square form
so ...... 37+1+n=2058
ans is 2020

Is this solution Helpfull? Yes (1) | No   

niki  (12 years ago)

if 2020 is not the answer thn how v'll solve..? pls reply..

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shivi vaish  (12 years ago)

not getting........pls explain well

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deepak srivastava  (12 years ago)

@ trishna
not getting your solution.

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Prateek  (12 years ago)

its a dummy ques.

Is this solution Helpfull? Yes | No (1)  

sandeep  (12 years ago)

plz expln it clearly plzzzzz plzzzzz

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