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

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(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

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$0$

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(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

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