let m=(82! /21). find the smallest positive integer x such that m≡x(mod 83)

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let m=(82! /21). find the smallest positive integer x such that m≡x(mod 83)

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Answer 1

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m= 881 21 as we know 83 is a prime and from willson theoren (P+) = 1 modo Where pre prime so putting p=83 so (83-1) = 1 mode 83 = 3x2 + 20 21= 1x20 #1 so = 21-20 = 21- (83- 3x21) 1=21-83+3x2) = 4x21 - 83 = 4x2 1 + (+) X 83 go 4 is inverse cef 2 in mo

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let m=(82! /21). find the smallest positive integer x such that m≡x(mod 83)

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let m=(82! /21). find the smallest positive integer x such that m≡x(mod 83)

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let m=(82! /21). find the smallest positive integer x such that m≡x(mod 83)