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

Answer

$1 \times _{10}1 =$ 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 $1.$
$⇒ a * 1 = 1 * a = a,  \forall\text{ a}\in\text{S}$
So, the identity element is $1.$
Also,
$3 \times _{10}7 = 1$
$3^{-1} = 7$

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