Latest Bank Exam Aptitude Question SOLUTION: The expression 2^6n - 4^2n, where n is the natural number is always divisble by ?

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(#M40034777) BANK EXAM Numerical Ability question numerical Keep an EYE

The expression 2^6n - 4^2n, where n is the natural number is always divisble by ?

Asked In Bank Exam shanmugapriya (10 years ago)

Solved RAKESH Read Solution (4)

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RAKESH  (10 years ago)

ans: 48

2^6n - 4^2n
= (64^n - 16^n)

(x^n - y^n) is always divisible by (x-y) for any natural no. n

so, (64^n - 16^n) is always divisble by (64-16)=48

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Rakesh Kumar Kohli  (10 years ago)

As (x-y)is a factor of (x^n - y^n) for any natural number
So (x^n - y^n) is always divisible by (x-y) for any natural number

Now The expression 2^6n - 4^2n = (2^6)^n-(2^4)^n where n is the natural number Hence the expression 2^6n - 4^2n is always divisible by 2^6 - 2^4 = 48

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RAM NARAYAN PATRO  (10 years ago)

always divisible by 48; because by assuming n=1.... we can proceed as
2^6-4^2=48
so if you put n=1,2,3,4...so on then you will get the numbers divisible by 48

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harish  (10 years ago)

ans is 48 [x^n-y^n is always divisable by x-y]

64^n-16^n =64-16=48

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