# Find Nth number in a sequence which is not a multiple of a given number

Given four integers A, N, L and R, the task is to find the N th number in a sequence of consecutive integers from L to R which is not a multiple of A. It is given that the sequence contain at least N numbers which are not divisible by A and the integer A is always greater than 1.

Examples:

Input: A = 2, N = 3, L = 1, R = 10
Output:
Explanation:
The sequence is 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. Here 5 is the third number which is not a multiple of 2 in the sequence.

Input: A = 3, N = 6, L = 4, R = 20
Output: 11
Explanation :
11 is the 6th number which is not a multiple of 3 in the sequence.

Naive Approach: The naive approach is to iterate over the range [L, R] in a loop to find the Nth number. The steps are:

1. Initialize the count of non-multiple number and current number to 0.
2. Iterate over the range [L, R] until the count of the non-multiple number is not equal to N.
3. Increment the count of the non-multiple number by 1, If the current number is not divisible by A.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach  #include  using namespace std;    // Function to find Nth number not a  // multiple of A in the range [L, R]  void count_no (int A, int N, int L, int R)  {            // To store the count      int count = 0;      int i = 0;        // To check all the nos in range      for(i = L; i < R + 1; i++)      {      if (i % A != 0)          count += 1;                if (count == N)          break;      }      cout << i;  }    // Driver code  int main()  {            // Given values of A, N, L, R      int A = 3, N = 6, L = 4, R = 20;            // Function Call      count_no (A, N, L, R);      return 0;  }    // This code is contributed by mohit kumar 29

## Java

 // Java program for the above approach  import java.util.*; import java.io.*;   class GFG{       // Function to find Nth number not a  // multiple of A in the range [L, R]  static void count_no (int A, int N,                        int L, int R) {        // To store the count      int count = 0;      int i = 0;        // To check all the nos in range      for(i = L; i < R + 1; i++)      {          if (i % A != 0)              count += 1;                if (count == N)              break;      }      System.out.println(i);  }    // Driver code  public static void main(String[] args)  {           // Given values of A, N, L, R      int A = 3, N = 6, L = 4, R = 20;        // Function call      count_no (A, N, L, R);  } }   // This code is contributed by sanjoy_62

## Python3

 # Python3 program for the above approach    # Function to find Nth number not a  # multiple of A in the range [L, R]  def count_no (A, N, L, R):        # To store the count      count = 0       # To check all the nos in range      for i in range ( L, R + 1 ):          if ( i % A != 0 ):              count += 1           if ( count == N ):              break     print ( i )        # Given values of A, N, L, R  A, N, L, R = 3, 6, 4, 20   # Function Call  count_no (A, N, L, R)

## C#

 // C# program for the above approach  using System;   class GFG{       // Function to find Nth number not a  // multiple of A in the range [L, R]  static void count_no (int A, int N,                        int L, int R) {            // To store the count      int count = 0;      int i = 0;        // To check all the nos in range      for(i = L; i < R + 1; i++)      {          if (i % A != 0)              count += 1;                if (count == N)              break;      }      Console.WriteLine(i);  }    // Driver code  public static void Main()  {           // Given values of A, N, L, R      int A = 3, N = 6, L = 4, R = 20;        // Function call      count_no (A, N, L, R);  } }   // This code is contributed by sanjoy_62

## Javascript

 

Output:

11

Time Complexity: O(R – L)
Auxiliary Space: O(1)

Efficient Approach:
The key observation is that there are A – 1 numbers that are not divisible by A in the range [1, A – 1]. Similarly, there are A – 1 numbers not divisible by A in range [A + 1, 2 * A – 1], [2 * A + 1, 3 * A – 1] and so on.
With the help of this observation, the Nth number which is not divisible by A will be:

To find the value in the range [ L, R ], we need to shift the origin from ‘0’ to ‘L – 1’, thus we can say that the Nth number which is not divisible by A in the range will be :

However there is an edge case, when the value of ( L â€“ 1 ) + N + floor( ( N â€“ 1 ) / ( A â€“ 1 ) ) itself turns out to be multiple of a ‘A’, in that case Nth number will be :

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include  using namespace std;   // Function to find Nth number // not a multiple of A in range [L, R] void countNo(int A, int N, int L, int R) {           // Calculate the Nth no     int ans = L - 1 + N + floor((N - 1) /                                  (A - 1));           // Check for the edge case     if (ans % A == 0)     {         ans = ans + 1;     }     cout << ans << endl; }   // Driver Code int main() {           // Input parameters      int A = 5, N = 10, L = 4, R = 20;           // Function Call     countNo(A, N, L, R);           return 0; }   // This code is contributed by avanitrachhadiya2155

## Java

 // Java program for the above approach import java.io.*;   class GFG  {     // Function to find Nth number   // not a multiple of A in range [L, R]   static void countNo(int A, int N, int L, int R)   {       // Calculate the Nth no     int ans = L - 1 + N + (int)Math.floor((N - 1) / (A - 1));       // Check for the edge case     if (ans % A == 0)     {       ans = ans + 1;     }     System.out.println(ans);       }     // Driver Code   public static void main (String[] args)    {       // Input parameters      int A = 5, N = 10, L = 4, R = 20;       // Function Call     countNo(A, N, L, R);       } }   //  This code is contributed by rag2127

## Python3

 # Python3 program for the above approach import math   # Function to find Nth number # not a multiple of A in range [L, R] def countNo (A, N, L, R):       # Calculate the Nth no     ans = L - 1 + N \           + math.floor( ( N - 1 ) / ( A - 1 ) )           # Check for the edge case     if ans % A == 0:         ans = ans + 1;     print(ans)   # Input parameters  A, N, L, R = 5, 10, 4, 20   # Function Call countNo(A, N, L, R)

## C#

 // C# program for the above approach using System; class GFG {     // Function to find Nth number   // not a multiple of A in range [L, R]   static void countNo(int A, int N, int L, int R)   {       // Calculate the Nth no     int ans = L - 1 + N + ((N - 1) / (A - 1));       // Check for the edge case     if (ans % A == 0)     {       ans = ans + 1;     }     Console.WriteLine(ans);   }     // Driver code   static void Main()    {       // Input parameters      int A = 5, N = 10, L = 4, R = 20;       // Function Call     countNo(A, N, L, R);   } }   // This code is contributed by divyesh072019.

## Javascript

 

Output:

16

Time Complexity: O(1)
Auxiliary Space: O(1)

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