Question

For the binary operation multiplication modulo 10 (×₁₀) defined on the set S = {1, 3, 7, 9}, write the inverse of 3.

Answer 1

Here,

1 ${\times }_{10}$ 1 = Remainder obtained by dividing 1 $\times$ 1 by 10
= 1

3 ${\times }_{10}$ 1 = Remainder obtained by dividing 3 $\times$ 1 by 10
= 3

7 ${\times }_{10}$ 3 = Remainder obtained by dividing 7 $\times$ 3 by 10
= 1

3 ${\times }_{10}$ 3 = Remainder obtained by dividing 3$\times$ 3 by 10
= 9

So, the composition table is as follows:

${\times }_{10}$	1	3	7	9
1	1	3	7	9
3	3	9	1	7
7	7	1	9	3
9	9	7	3	1

We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.

These two intersect at 1.
$\Rightarrow a*1=1*a=a,\forall a\in S$
So, the identity element is 1.

Also,
3 ${\times }_{10}$ 7 = 1
3^-1 = 7