For the binary operation x₁₀ on set S = {1, 3, 7, 9}, find the inverse of e3.

A composition table consists of elements which are a result of operation on the set elements.

Here we have the operation, a x₁₀b = remainder of ab divided by 10 where a, b S.

For bS to be an inverse of aS, a x₁₀b = e, where e is the identity element.

We know for multiplication operation we have the identity element as 1.

So e = 1.

For a = 3,

3 x₁₀ (inverse of 3) = 1

From the table above, 3 x₁₀ 7 = 1

Hence we can conclude that ‘inverse of 3’ must be 7.