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| # 17 x = 10 mod 50
# xgcd function from exercise 1.14
xgcd(17,50)
> (1, 3, -1)
# 3 is the modular inverse of 17 mod 50
Mod(3*17,50)
> 1
x=Mod(3*10,50); x
> 30
xgcd(35,50)
> (5, 3, -2)
# 35 is not invertible mod 50
# but 5=gcd(35,50) divides 10
# Thus the equation is transformed into 7 x = 2 mod 10
xgcd(7,10)
> (1, 3, -2)
# 3 is the modular inverse of 7 mod 10
Mod(3*7,10)
> 1
x=Mod(3*2,10);x
> 6
# 35 is not invertible mod 50
# but 5=gcd(35,50) does not divide 11
# There is no solution to 35 x = 11 mod 50
|
