The remainder when (11)^{1011} + (1011)^{11} | vedclass

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(D) We need to find the remainder when $(11)^{1011} + (1011)^{11}$ is divided by $9$.First,observe that $11 \equiv 2 \pmod{9}$ and $1011 = 9 	imes 11

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) We need to find the remainder when $(11)^{1011} + (1011)^{11}$ is divided by $9$.First,observe that $11 \equiv 2 \pmod{9}$ and $1011 = 9 	imes 112 + 3$,so $1011 \equiv 3 \pmod{9}$.Thus,$(11)^{1011} + (1011)^{11} \equiv 2^{1011} + 3^{11} \pmod{9}$.For $2^{1011} \pmod{9}$:$2^6 = 64 \equiv 1 \pmod{9}$.Since $1011 = 6 	imes 168 + 3$,we have $2^{1011} = (2^6)^{168} 	imes 2^3 \equiv 1^{168} 	imes 8 \equiv 8 \pmod{9}$.For $3^{11} \pmod{9}$:$3^2 = 9 \equiv 0 \pmod{9}$,so $3^{11} = 3^2 	imes 3^9 = 9 	imes 3^9 \equiv 0 \pmod{9}$.Therefore,$(11)^{1011} + (1011)^{11} \equiv 8 + 0 \equiv 8 \pmod{9}$.The remainder is $8$.

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

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