Question

Suppose, the remainder obtained while dividing x by 61 is 2. What is the remainder obtained while dividing X^7 by 61?​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{The remainder obtained when dividing x by}\;$$\textsf{61 is 2}$

$\underline{\textbf{To find:}}$

$\textsf{The remainder obtained when}\;\mathsf{x^7\;is}$

$\mathsf{divided\;by\;61}$

$\underline{\textbf{Solution:}}$

$\textsf{We apply binomial theorem to solve this problem}$

$\underline{\textbf{Binomial theorem:}}$

$\textsf{If 'n' is a positive integer, then}$

$\boxed{\mathsf{(a+b)^n=n_{C_0}\,a^n+n_{C_1}\,a^{n-1}\,b+n_{C_2}\,a^{n-2}\,b^2+\;.\;.\;.\;.+n_{C_{n-1}}\,a^1\,b^{n-1}+n_{C_n}\,b^n}}$

$\textsf{when  x is divided by 61 the remainder is 2}$

$\textsf{By Division algorithm,}$

$\mathsf{x=61\,k+2}$

$\mathsf{Now,}$

$\mathsf{x^7}$

$\mathsf{=(61\,k+2)^7}$

$\mathsf{=7_{C_0}\,(61k)^7+7_{C_1}\,(61k)^6\,2\,b+7_{C_2}\,(61k)^5\,2^2+\;.\;.\;.\;.+7_{C_6}\,(61k)^1\,2^6+7_{C_7}\,2^7}$

$\mathsf{=(Multiple\;of\;61)+7_{C_7}(128)}$

$\mathsf{=(Multiple\;of\;61)+1(128)}$

$\mathsf{=(Multiple\;of\;61)+(2\,\times\,61)+6}$

$\mathsf{=(Multiple\;of\;61)+6}$

$\implies\textsf{The remainder when}\,\mathsf{x^7\;is\;divided\;by\;61\;is\;6}$

$\underline{\textbf{Find more:}}$

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