Question

What is the remainder when ${82^{361}} + {83^{361}} + {84^{361}} + {85^{361}} + {86^{361}}$ is divided by $7$?

Answer 1

Hint: In the above question, first we will write the above equation in such a form so that we can find the multiple of $7$ so that we can find the remainder. Then we will add all the remainders to find the final answer.

Complete step by step answer:
In the above question, we have
${82^{361}} + {83^{361}} + {84^{361}} + {85^{361}} + {86^{361}}$
We can also write it as,
$ \Rightarrow {\left( {84 - 2} \right)^{361}} + {\left( {84 - 1} \right)^{361}} + {\left( {84} \right)^{361}} + {\left( {84 + 1} \right)^{361}} + {\left( {84 + 2} \right)^{361}}$
We know that $84$ is a multiple of $7$.
Since \[84\] is a multiple of $7$ then the remainder will be when ${\left( {84 - 2} \right)^{361}} + {\left( {84 - 1} \right)^{361}} + {\left( {84} \right)^{361}} + {\left( {84 + 1} \right)^{361}} + {\left( {84 + 2} \right)^{361}}$ is divided by $7$ is ${\left( { - 2} \right)^{361}} + {\left( { - 1} \right)^{361}} + {\left( 1 \right)^{361}} + {\left( 2 \right)^{361}}$.
Now, we have
${\left( { - 2} \right)^{361}} + {\left( { - 1} \right)^{361}} + {\left( 1 \right)^{361}} + {\left( 2 \right)^{361}}$
We can also write it as,
${\left( { - 1} \right)^{361}}{\left( 2 \right)^{361}} + {\left( { - 1} \right)^{361}}{\left( 1 \right)^{361}} + {\left( 1 \right)^{361}} + {\left( 2 \right)^{361}}$
Here the powers of $ - 1$ are $361$ and $361$ is an odd number. So, we will get $ - 1$ as the final result after solving the powers.
So,
$ \Rightarrow - 1{\left( 2 \right)^{361}} + - 1{\left( 1 \right)^{361}} + {\left( 1 \right)^{361}} + {\left( 2 \right)^{361}}$
On adding these terms, we will get
$ \Rightarrow 0$

Hence, the remainder when ${82^{361}} + {83^{361}} + {84^{361}} + {85^{361}} + {86^{361}}$ is divided by $7$ is $0$.

Note: Here our answer comes out to be zero. But if it was not zero then, first we have to find the sum of all the numbers and then we have to divide it by seven. We can also find the remainder by taking out the large numbers from the brackets.