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The remainder of (3)^67! divided by 80 is :

10
21
32
43

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Since, 3^{4} = 81 gives remainder 1 on divided by 80 \\ So, \frac{3^{4 n}}{80}, gives remainder 1. \\ Thus, \frac{3^{67!}}{80}, will also give the remainder as 1 \\ Since, 67! = 4 n for a positive integer n \end{aligned}

$\eqalign{ & {\text{Since}},\,{3^4} = 81\,{\text{gives}}\,{\text{remainder}}\,{\text{1}}\,{\text{on}}\,{\text{divided}}\,{\text{by}}\,80 \cr & {\text{So}},\,\frac{{{3^{4n}}}}{{80}},\,{\text{gives}}\,{\text{remainder}}\,1. \cr & {\text{Thus}},\,\frac{{{3^{67!}}}}{{80}},\,{\text{will}}\,{\text{also}}\,{\text{give}}\,{\text{the}}\,{\text{remainder}}\,{\text{as}}\,1 \cr & {\text{Since}},\,67! = 4n\,{\text{for}}\,{\text{a}}\,{\text{positive}}\,{\text{integer}}\,n \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Since, 3^{4} = 81 gives remainder 1 on divided by 80 \\ So, \frac{3^{4 n}}{80}, gives remainder 1. \\ Thus, \frac{3^{67!}}{80}, will also give the remainder as 1 \\ Since, 67! = 4 n for a positive integer n \end{aligned}

$\eqalign{ & {\text{Since}},\,{3^4} = 81\,{\text{gives}}\,{\text{remainder}}\,{\text{1}}\,{\text{on}}\,{\text{divided}}\,{\text{by}}\,80 \cr & {\text{So}},\,\frac{{{3^{4n}}}}{{80}},\,{\text{gives}}\,{\text{remainder}}\,1. \cr & {\text{Thus}},\,\frac{{{3^{67!}}}}{{80}},\,{\text{will}}\,{\text{also}}\,{\text{give}}\,{\text{the}}\,{\text{remainder}}\,{\text{as}}\,1 \cr & {\text{Since}},\,67! = 4n\,{\text{for}}\,{\text{a}}\,{\text{positive}}\,{\text{integer}}\,n \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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