Find the Nth natural number which is not divisible by A
Given two integers A and N, our task is to find the Nth natural number which is not divisible by A.
Input: A = 4, N = 12
The series starting from 1 excluding the multiples of A would be 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17 and the 12th term which is not divisible by 4 is 15.
Input: A = 3, N = 20
The series starting from 1 excluding the multiples of A would be 1, 2, 4, 5, 7, 8, 10, 11 and so on and the Nth number which is not divisible by 3 is 29.
To solve the problem mentioned above we have to observe that there is a gap after every (A – 1) integer as the multiples of A are skipped. To find that the number N lies in which set we divide N by (A – 1), and store it in a variable lets say quotient. Now multiplying A with that variable and adding the remainder to it we will get the resultant answer.
If A = 4, N = 7: quotient = 7 / 3 = 2 remainder = 7 % 3 = 1 So the answer is (A * quotient) + remainder = 4 * 2 + 1 = 9
But if the remainder is 0 it means it is the last element in the set. Lets us understand it by an example.
If A = 4, N = 6: quotient = 6 / 3 = 2 remainder = 6 % 3 = 0 So the answer is (A * quotient) - 1 = 4 * 2 - 1 = 7
Below is the implementation of the above approach:
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