This can be solved by Chinese remainder theorem, but as the common difference is constant, it is a special case of Chinese remainder concept.

Required number of the set is calculated by the LCM of (2, 3, 4, 5, 6) - (common difference)

In this case, common difference = (2-1) = (3-2) = (4-3) = (5- 4) = (6 - 5) = 1.

All integers of the set will be given by (60n - 1) ; If n = 1, (60 - 1) = 59; If n = 2,((60*2) - 1) = 119; Since range of the set A is between 0 and 100, hence there will exist only one number i.e., 59.