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

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

Question

(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

Similar Questions

The remainder,when $7^{103}$ is divided by $17$ is $..........$.

The sum $1! + 4! + 7! + 10! + 12! + 13! + 15! + 16! + 17!$ is divisible by

The least positive integer greater than $1$ that divides $49^n + 16n - 1$ for all positive integers $n$ is

If $11^{12}-11^2=k(5 \times 10^9+6 \times 10^9+33 \times 10^8+110 \times 10^7+\ldots+33)$,then $k=$

The remainder when $((64)^{(64)})^{(64)}$ is divided by $7$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module