Find the inverse of 35 modulo 11 by using extended Euclidean Algorithm
step by step solution
"35=3\\times11+2...(1)\\\\ \n11=5\\times2+1...(2)\\\\\n2=2\\times1+0\\\\"
Now,
"1=11-5\\times2\\ \\text{From (2)}\\\\\n1= 11-5\\times (35-3\\times11) \\ \\text{From (1)}\\\\\n1= 11-5\\times35+15\\times11\\\\\n1=-5\\times35+16\\times11\\\\\n1(\\text{mod 11})=6\\times35+16\\times11\\\\\n1(\\text{mod 11})=6\\times35"
"6\\equiv 35^{-1} \\text{mod 11}"
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