Question

Let +₆ (addition modulo 6) be a binary operation on S = {0, 1, 2, 3, 4, 5}. Write the value of \[2 +_6 4^{- 1} +_6 3^{- 1} .\]

Answer 1

Here,

1 \[+_6\] 1 = Remainder obtained by dividing 1 + 1 by 6
           = 2

3 \[+_6\] 4 = Remainder obtained by dividing 3 + 4 by 6
            = 1

4 \[+_6\] 5 = Remainder obtained by dividing 4 + 5 by 6
           = 3 

So, the composition table is as follows:

+6	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

We observe that the first row of the composition table coincides with the the top-most row and the first column coincides with the left-most column.
These two intersect at 0.

\[\Rightarrow a +_6 0 = 0 +_6 a = a, \forall a \in S\] 

So, 0 is the identity element. 

From the table,

\[4 +_6 2 = 0 \]
\[ \Rightarrow 4^{- 1} = 2\]
\[3 +_6 3 = 0 \]
\[ \Rightarrow 3^{- 1} = 3\]
\[\text{Now},\]
\[2 +_6 4^{- 1} +_6 3^{- 1} = 2 +_6 2 +_6 3\]
\[ = 4 +_6 3 \]
\[ = 1\]