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 = 12Output:15Explanation:

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 = 20Output:29Explanation:

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++ 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 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 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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`// C# code to find the Nth number which ` `// is not divisible by A from the series ` `using` `System; `
` ` `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 `
` ` `Console.WriteLine(a); `
`} ` ` ` `// Driver code ` `public` `static` `void` `Main(String[] args) `
`{ ` ` ` `int` `A = 4; `
` ` `int` `N = 6; `
` ` ` ` `findNum(A, N); `
`} ` `} ` ` ` `// This code is contributed by amal kumar choubey ` |

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**Output:**

7

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