Answer:
a) the additive inverse of 4 mod 11 is 7.
b) The multiplicative inverse of 72 mod 11 is 2.
c) x = 5
Step-by-step explanation:
The additive inverse of a number a in mod b is a number c such that a + c = 0 (mod b)
We need to find a number b such that 4 + b = 0 (mod 11)
4 + b = 0 (mod 11)
4 + b = 11k (mod 11) (if it's 0 (mod 11) then it's multiple of 11 and can be written as 11k
b = 11k - 4 (mod 11)
b = - 4 (mod 11)
b = 7 (mod 11)
Therefore, the additive inverse of 4 mod 11 is 7.
b) In mod 11 arithmetic what is the multiplicative inverse of 72?
The multiplicative inverse of a number a in mod b is a number c such that
ac = 1 mod b
First we see that 72 is larger than 11, 72/11 = 6(11) + 6.
Therefore 72 = 6 (mod 11)
We need to find a number c such that 6c = 1 (mod 11)
This means we need to find a number c such that:
6c + 11a = 1 and this c will be the multiplicative inverse of 6.
We divide 11 by 6 and we get
11 = 6(1) + 5
Now we divide 6 by 5 and we get:
6 = 5(1) + 1
Therefore
1 = 6 - 5(1)
But we knew that 5 is 11 - 6 (we are going to work back the other way)
1 = 6 - (11-6)(1)
1 = 6 - 11(1) +6(1)
1 = -11(1) + 2(6)
1= 2(6) -11(1) = 6(2) +11(-1)
Thus c = 2 and this is the multiplicative inverse of 72.
c) Show all steps in solving the congruency equation in mod 11. 7x + 4 =115
7x + 4 = 115 (mod 11)
We divide 115/11 and we get 115 = 11(10) + 5.
So we can rewrite the equation to: 7x + 4 = 5 (mod 11)
7x + 4 = 5 (mod 11)
7x = 5 - 4 = 1 (mod 11)
7x = 1 (mod 11)
Therefore we need to find the multiplicative inverse of 7 mod 11.
7 x 1 = 7 (mod 11)
7 x 2 = 14 = 3 (mod 11)
7 x 3 = 21 = 10 (mod 11)
7 x 4 = 28 = 6 (mod 11)
7 x 5 = 35 = 2 (mod 11)
7 x 6 = 42 = 9 (mod 11)
7 x 7 = 49 = 5 (mod 11)
7 x 8 = 56 = 1 (mod 11)
7 x 9 = 63 = 8 (mod 11)
7 x 10 = 70 = 4 (mod 11)
Thus, the multiplicative inverse of 7 mod 11 is 8. and the solution of 7x + 4 = 115 (mod 11) is x = 8
