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
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<bits/stdc++.h>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 Codeint 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 approachimport 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 Codepublic 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 approachimport 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 = 15if isDNum(n): print("Yes")else: print("No") |
C#
// C# implementation for the // above approachusing 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 Codepublic 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 |
Javascript
<script>// Javascript implementation for the // above approach // Function to find the N-th // icosikaipentagon number function isDNum( n) { // Number should be // greater than 3 if (n < 4) return false; let numerator = 0, hcf = 0; // Check every k in range 2 to n-1 for ( k = 2; k <= n; k++) { numerator = parseInt( (Math.pow(k, n - 2) - k)); hcf = __gcd(n, k); } // Condition for D-Number if (hcf == 1 && (numerator % n) != 0) return false; return true; } function __gcd( a, b) { return b == 0 ? a : __gcd(b, a % b); } // Driver Code let n = 15; let a = isDNum(n); if (a) document.write("Yes"); else document.write("No");// This code contributed by Rajput-Ji </script> |
Output:
Yes
Time Complexity: O(1)
Reference: OEIS
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