1. (a) Draw addition⊕ and multiplication ⊗ tables for arithmetic modulo 6.
(b) Use your tables to solve these equations:
(i) 5 ⊗n=1 (ii) 2⊕n=0 (iii) 2⊕(3 ⊕n)=1
1a ⊕ 0 1 2 3 4 5
0 0 1 2 3 4 5
1 1 2 3 4 5 0
2 2 3 4 5 0 1
3 3 4 5 0 1 2
4 4 5 0 1 2 3
5 5 0 1 2 3 4
Addition tables in modulo 6
⊗ 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 1 2 3 4 5
2 0 2 4 0 2 4
3 0 3 0 3 0 3
4 0 4 2 0 4 2
5 0 5 4 3 2 1
Multiplication table in modulo 6
bi 5⊗n=1
From tables n=5
ii. 2⊕n=0
From table n=4
iii. 2⊕(3⊕n)=1
From tables (3⊕n) is 5
(3⊕n)=5
From tables n=2
Note: between each row and column in the tables there are lines which I couldn't put