# 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.

**Examples:**

Input:A = 4, N = 12

Output:15

Explanation:

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

Output:29

Explanation:

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.

Approach:

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:

## C++

`// C++ code to Find the Nth number which ` `// is not divisible by A from the series ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `void` `findNum(` `int` `n, ` `int` `k) ` `{ ` ` ` ` ` `// Find the quotient and the remainder ` ` ` `// when k is divided by n-1 ` ` ` `int` `q = k / (n - 1); ` ` ` `int` `r = k % (n - 1); ` ` ` `int` `a; ` ` ` ` ` `// If the remainder is not 0 ` ` ` `// multiply n by q and subtract 1 ` ` ` `// if remainder is 0 then ` ` ` `// multiply n by q ` ` ` `// and add the remainder ` ` ` `if` `(r != 0) ` ` ` `a = (n * q) + r; ` ` ` `else` ` ` `a = (n * q) - 1; ` ` ` ` ` `// Print the answer ` ` ` `cout << a; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `A = 4, N = 6; ` ` ` ` ` `findNum(A, N); ` ` ` `return` `0; ` `} ` ` ` `// This code is contributed by PratikBasu ` |

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## Java

`// Java code to Find the Nth number which ` `// is not divisible by A from the series ` `class` `GFG { ` ` ` `static` `void` `findNum(` `int` `n, ` `int` `k) ` `{ ` ` ` ` ` `// Find the quotient and the remainder ` ` ` `// when k is divided by n-1 ` ` ` `int` `q = k / (n - ` `1` `); ` ` ` `int` `r = k % (n - ` `1` `); ` ` ` `int` `a = ` `0` `; ` ` ` ` ` `// If the remainder is not 0 ` ` ` `// multiply n by q and subtract 1 ` ` ` `// if remainder is 0 then ` ` ` `// multiply n by q ` ` ` `// and add the remainder ` ` ` `if` `(r != ` `0` `) ` ` ` `a = (n * q) + r; ` ` ` `else` ` ` `a = (n * q) - ` `1` `; ` ` ` ` ` `// Print the answer ` ` ` `System.out.println(a); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `A = ` `4` `; ` ` ` `int` `N = ` `6` `; ` ` ` ` ` `findNum(A, N); ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar ` |

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## Python3

`# Python code to Find the Nth number which ` `# is not divisible by A from the series ` ` ` `def` `findNum(n, k): ` ` ` ` ` `# Find the quotient and the remainder ` ` ` `# when k is divided by n-1 ` ` ` `q ` `=` `k` `/` `/` `(n` `-` `1` `) ` ` ` `r ` `=` `k ` `%` `(n` `-` `1` `) ` ` ` ` ` `# if the remainder is not 0 ` ` ` `# multiply n by q and subtract 1 ` ` ` ` ` `# if remainder is 0 then ` ` ` `# multiply n by q ` ` ` `# and add the remainder ` ` ` `if` `(r !` `=` `0` `): ` ` ` `a ` `=` `(n ` `*` `q)` `+` `r ` ` ` `else` `: ` ` ` `a ` `=` `(n ` `*` `q)` `-` `1` ` ` ` ` `# print the answer ` ` ` `print` `(a) ` ` ` `# driver code ` `A ` `=` `4` `N ` `=` `6` `findNum(A, N) ` |

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**Output:**

7

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