Question

If ${(2021)^{3762}}$ is divided by $17$ ,then the remainder is

Answer 1

Hint: It’s very simple to do this type of question only you have to simplify it by breaking the system. for example break the numbers into divisible by the same one and by using such a common method we can easily complete this type of question.

Complete step by step Solution:
Firstly as given in the question, we should take ${(2021)^{3762}}$ and break it into its nearest in which it can be easily divisible by $17$.
${(2021)^{3762}}$ we can also write it as,
${(2021)^{3762}}$ $ = {(2023 - 2)^{3762}} = $ (Here $2023$ is a multiple of 17 we can also write it as)
$ = {(2023 - 2)^{3762}} = $ multiple of $17 + {2^{3762}}$
From which we can also write is the form of,
$ = 17\lambda + {2^2}{({2^4})^{940}}$
As it is expended in this from now taking further solution we get,
$ = 17\lambda + 4{(17 - 1)^{940}}$
As we can re-change the power into exponent and write the whole equation as,
$ = 17\lambda + 4(17\mu + 1)$
Here from this equation, we get that,
$ = 17k + 4$
From the above equation, we know that here $(k \in 1)$ from which get that,
Hence, we get the answer as 4.
Therefore, the reminder in ${(2021)^{3762}}$ with divisible by $17$ is \[4\] .

Note: In doing this type of question we always need to know the divisible by term in this matter. Whereas, expansions and powers are also very useful in this and always have to focus on the equation term a easy and useful part is there which is to use a given term in every question by which you easily know what the question is said about in it and get all values in a place for easy calculations.