If (11)^{27} + (21)^{27} is divided by 16,the | vedclass

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(A) We know that for any odd positive integer $n$,$a^n + b^n$ is divisible by $(a + b)$.Here,$n = 27$,which is an odd number.Therefore,$(11)^{27} + (2

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$0$

Vedclass · Answer

(A) We know that for any odd positive integer $n$,$a^n + b^n$ is divisible by $(a + b)$.Here,$n = 27$,which is an odd number.Therefore,$(11)^{27} + (21)^{27}$ is divisible by $(11 + 21) = 32$.Since $32$ is a multiple of $16$ (i.e.,$32 = 2 	imes 16$),any number divisible by $32$ must also be divisible by $16$.Thus,$(11)^{27} + (21)^{27}$ is divisible by $16$.When a number is perfectly divisible by the divisor,the remainder is $0$.

If $(11)^{27} + (21)^{27}$ is divided by $16$,the remainder is:

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