Answer to Question #264361 in Discrete Mathematics for Helen

Question #264361

Find the inverse of 35 modulo 11 by using extended Euclidean Algorithm


1
Expert's answer
2021-11-15T16:41:09-0500

Let us find the inverse of 35 modulo 11 by using extended Euclidean Algorithm.

Taking into account that

35=113+235=11\cdot 3+2\\ and 11=25+1,11=2\cdot5+1,

we conclude that

1=1125=11(35113)5=1116+35(5).1=11-2\cdot 5=11-(35-11\cdot 3)5=11\cdot 16+35(-5).

It follows that

1116+35(5)=1 mod 11,11\cdot 16+35(-5)=1\ mod\ 11,

and hence

35(5)=1 mod 11.35(-5)=1\ mod\ 11.

Since 5 mod 11=6 mod 11,-5\ mod\ 11=6\ mod\ 11, we conclude that the inverse of 35 modulo 11 is 6.


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