The remainder,when (15^{23} + 23^{23}) is div | vedclass

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

Question

(C) We need to find the remainder when $(15^{23} + 23^{23})$ is divided by $19$.Note that $15 \equiv -4 \pmod{19}$ and $23 \equiv 4 \pmod{19}$.Therefo

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) We need to find the remainder when $(15^{23} + 23^{23})$ is divided by $19$.Note that $15 \equiv -4 \pmod{19}$ and $23 \equiv 4 \pmod{19}$.Therefore,$15^{23} + 23^{23} \equiv (-4)^{23} + 4^{23} \pmod{19}$.Since $23$ is an odd integer,$(-4)^{23} = -4^{23}$.Thus,$15^{23} + 23^{23} \equiv -4^{23} + 4^{23} \equiv 0 \pmod{19}$.Hence,the remainder is $0$.

The remainder,when $(15^{23} + 23^{23})$ is divided by $19$,is

Similar Questions

The remainder when $(11)^{1011} + (1011)^{11}$ is divided by $9$ is

If $a, b$ and $n$ are natural numbers,then $a^{2n-1} + b^{2n-1}$ is divisible by

If $n > 1$ is an integer and $x \neq 0$, then $(1+x)^{n}-nx-1$ is divisible by

If the greatest divisor of $30 \cdot 5^{2n} + 4 \cdot 2^{3n}$ is $p, \forall n \in N$ and that of $2^{2n+1} - 6n - 2$ is $q, \forall n \in N$, then $p+q=$

The remainder obtained when $5^{99}$ is divided by $13$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module