Find the inverse of 35 modulo 11 by using extended Euclidean Algorithm
Let us find the inverse of 35 modulo 11 by using extended Euclidean Algorithm.
Taking into account that
"35=11\\cdot 3+2\\\\" and "11=2\\cdot5+1,"
we conclude that
"1=11-2\\cdot 5=11-(35-11\\cdot 3)5=11\\cdot 16+35(-5)."
It follows that
"11\\cdot 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.
Comments
Leave a comment