# D Numbers

D Number is a number N > 3 such that N divides k^(n-2)-k for all k with gcd(k, n) = 1, 1<k<n.

9, 15, 21, 33, 39, 51, 57, 63, 69, 87, 93….

### Check if a number is a D-Number

Given a number N, the task is to check if N is an D Number or not. If N is an D Number then print “Yes” else print “No”.

Examples:

Input: N = 9
Output: Yes
Explanation:
9 is a D-number since it divides all the numbers
2^7-2, 4^7-4, 5^7-5, 7^7-7 and 8^7-8, and
2, 4, 5, 7, 8 are relatively prime to n.

Input: N = 16
Output: No

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Since D Number is a number N > 3 such that N divides k^(n-2)-k for all k with gcd(k, n) = 1, 1<k<n. So in a loop of k from 2 to n-1, we will check if gcd of N and k is 1 or not.If it's 1 then we will check if k^(n-2)-k is divisible by N or not.If not divisible we will return false.At last we will return true.

Below is the implementation of the above approach:

## C++

 // C++ implementation  // for the above approach #include using namespace std;    // Function to find the N-th  // icosikaipentagon number  int isDNum(int n) {     // number should be      // greater than 3     if (n < 4)         return false;            int numerator, hcf;            // Check every k in range 2 to n-1     for (int k = 2; k <= n; k++)      {         numerator = pow(k, n - 2) - k;         hcf = __gcd(n, k);     }            // condition for D-Number     if (hcf == 1 && (numerator % n) != 0)         return false;            return true; }    // Driver Code int main()  {     int n = 15;     int a = isDNum(n);     if (a)         cout << "Yes";     else         cout << "No"; }    // This code is contributed by Ritik Bansal

## Java

 // Java implementation for the  // above approach import java.util.*;    class GFG{    // Function to find the N-th  // icosikaipentagon number  static boolean isDNum(int n) {            // Number should be      // greater than 3     if (n < 4)         return false;            int numerator = 0, hcf = 0;            // Check every k in range 2 to n-1     for(int k = 2; k <= n; k++)      {        numerator = (int)(Math.pow(k, n - 2) - k);        hcf = __gcd(n, k);     }            // Condition for D-Number     if (hcf == 1 && (numerator % n) != 0)         return false;            return true; }    static int __gcd(int a, int b)  {      return b == 0 ? a : __gcd(b, a % b);      }     // Driver Code public static void main(String[] args)  {     int n = 15;     boolean a = isDNum(n);            if (a)         System.out.print("Yes");     else         System.out.print("No"); } }    // This code is contributed by Amit Katiyar

## Python3

 # Python3 implementation  # for the above approach    import math       # Function to find the N-th  # icosikaipentagon number  def isDNum(n):      # number should be          # greater than 3     if n < 4:         return False        # Check every k in range 2 to n-1     for k in range(2, n):         numerator = pow(k, n - 2) - k         hcf = math.gcd(n, k)            # condition for D-Number         if(hcf ==1 and (numerator % n) != 0):             return False     return True    # Driver code  n = 15 if isDNum(n):     print("Yes") else:     print("No")

## C#

 // C# implementation for the  // above approach using System; class GFG{    // Function to find the N-th  // icosikaipentagon number  static bool isDNum(int n) {            // Number should be      // greater than 3     if (n < 4)         return false;            int numerator = 0, hcf = 0;            // Check every k in range 2 to n-1     for(int k = 2; k <= n; k++)      {         numerator = (int)(Math.Pow(k, n - 2) - k);         hcf = __gcd(n, k);     }            // Condition for D-Number     if (hcf == 1 && (numerator % n) != 0)         return false;            return true; }    static int __gcd(int a, int b)  {      return b == 0 ? a : __gcd(b, a % b);      }     // Driver Code public static void Main(String[] args)  {     int n = 15;     bool a = isDNum(n);            if (a)         Console.Write("Yes");     else         Console.Write("No"); } }    // This code contributed by Princi Singh

Output:

Yes

Time Complexity: O(1)
Reference: OEIS

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