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.


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