Let us find the inverse of 35 modulo 11 by using extended Euclidean Algorithm.
Taking into account that
35=11⋅3+2 and 11=2⋅5+1,
we conclude that
1=11−2⋅5=11−(35−11⋅3)5=11⋅16+35(−5).
It follows that
11⋅16+35(−5)=1 mod 11,
and hence
35(−5)=1 mod 11.
Since −5 mod 11=6 mod 11, we conclude that the inverse of 35 modulo 11 is 6.
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