Question

A is a set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively. How many integers between 0 and 100 belong to set A?

• A. 1
• B. 2
• C. None of these

Answer 1

The correct option is A 1

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, number will be (60 - 1) = 59. If n = 2, number will be 60$\times$2 - 1 = 119.
Since range of the set A is between 0 and 100, hence there will exist only one number i.e. 59.