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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$0$

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:

Similar Questions

For all positive integers $k$, if the greatest divisor of $25^k+12k-1$ is $d$, then $4\sqrt{d}=$

Statement $-1$: For every natural number $n$,$(n + 1)^7 - n^7 - 1$ is divisible by $7$.
Statement $-2$: For every natural number $n$,$n^7 - n$ is divisible by $7$.

The expression $x(x^{n-1} - na^{n-1}) + a^n(n-1)$ is divisible by $(x-a)^2$ for:

The expression $(1 + x)^n - nx - 1$ is divisible by (where $n \in N$ and $n > 1$):

The remainder,when $3^{2003}$ is divided by $28$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module