Question #264361

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


Expert's answer

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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