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

• Difficulty Level : Medium
• Last Updated : 20 Mar, 2023

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 – 1 ) + N + floor( ( N – 1 ) / ( A – 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)

#### Another Approach:

1. We initialize n to the position of the required number in the sequence and x to the given number which should not be a multiple of the required number. In this example, we have taken n as 10 and x as 3.
2. We initialize a variable count to 0, which will keep track of the number of non-multiples encountered so far.
3. We use a for loop that starts from 1 and continues until we find the nth non-multiple of x. Inside the for loop, we check whether the current number i is a multiple of x.
4. If i is not a multiple of x, we set the variable num to the value of i and increment the count variable.
5. When count becomes equal to n, we exit the for loop and print the value of num as the nth number in the sequence which is not a multiple of x.

## C++

 #include  using namespace std; int main(){     int n = 10; // for example     int x = 3; // for example     int count = 0;     int i, num;       for (i = 1; count < n; i++) {         if (i % x != 0) {             num = i;             count++;         }     }     cout<< "The " << n << "th number in the sequence which is not a multiple of "       << x << " is " << num << endl;     return 0; }

## C

 #include    int main() {     int n = 10; // for example     int x = 3; // for example     int count = 0;     int i, num;       for (i = 1; count < n; i++) {         if (i % x != 0) {             num = i;             count++;         }     }       printf("The %dth number in the sequence which is not a multiple of %d is %d\n", n, x, num);       return 0; }

Output

The 10th number in the sequence which is not a multiple of 3 is 14

Time Complexity: O(N), where N is the position of the required number in the sequence.
Space Complexity: O(1), as we are not using any extra space.

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