Find out the prime numbers in the form of A+nB or B+nA
Given two integers A and B and an integer N. The task is to find out N prime numbers of the form A + nB or B + nA( n=1, 2, 3…). If it is not possible, print -1.
Examples:
Input: A = 3, B = 5, N = 4
Output: 13, 11, 17, 23
Explanation:
13 (3+2*5)
11 (5+2*3)
17 (5+4*3)
23 (3+4*5)Input: A = 4, B = 6, N = 4
Output: -1
Approach:
Since A + nB to be a prime one thing is sure that there should not present any common factor between A and B, means A and B should be co-prime.
The best and most efficient approach will be to use Dirichlet’s Theorem.
Dirichlet’s Theorem says that if a and b are relatively prime positive integers, then the arithmetic sequence a, a+b, a+2b, a+3b…contains infinitely many primes.
So firstly check if A and B are co-prime.
If A and B are co-prime then check A+nB and B+nA for their primality where n=1, 2, 3…. Print the first N prime numbers among them.
Below is the implementation of the above approach:
C++
// C++ implementation of // the above approach #include <bits/stdc++.h> using namespace std; // Utility function to check // whether two numbers is // co-prime or not int coprime(int a, int b) { if (__gcd(a, b) == 1) return true; else return false; } // Utility function to check // whether a number is prime // or not bool isPrime(int n) { // Corner case if (n <= 1) return false; if (n == 2 or n == 3) return true; // Check from 2 to sqrt(n) for (int i = 2; i * i <= n; i++) if (n % i == 0) return false; return true; } // finding the Prime numbers void findNumbers(int a, int b, int n) { bool possible = true; // Checking whether given // numbers are co-prime // or not if (!coprime(a, b)) possible = false; int c1 = 1; int c2 = 1; int num1, num2; // To store the N primes set<int> st; // If 'possible' is true if (possible) { // Printing n numbers // of prime while ((int)st.size() != n) { // checking the form of a+nb num1 = a + (c1 * b); if (isPrime(num1)) { st.insert(num1); } c1++; // Checking the form of b+na num2 = b + (c2 * a); if (isPrime(num2)) { st.insert(num2); } c2++; } for (int i : st) cout << i << " "; } // If 'possible' is false // return -1 else cout << "-1"; } // Driver Code int main() { int a = 3; int b = 5; int n = 4; findNumbers(a, b, n); return 0; } |
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Java
// Java implementation of the approach import java.util.*; class GFG { static int __gcd(int a, int b) { if (b == 0) return a; return __gcd(b, a % b); } // Utility function to check // whether two numbers is // co-prime or not static boolean coprime(int a, int b) { if (__gcd(a, b) == 1) return true; else return false; } // Utility function to check // whether a number is prime // or not static boolean isPrime(int n) { // Corner case if (n <= 1) return false; if (n == 2 || n == 3) return true; // Check from 2 to sqrt(n) for (int i = 2; i * i <= n; i++) if (n % i == 0) return false; return true; } // finding the Prime numbers static void findNumbers(int a, int b, int n) { boolean possible = true; // Checking whether given // numbers are co-prime // or not if (!coprime(a, b)) possible = false; int c1 = 1; int c2 = 1; int num1, num2; // To store the N primes HashSet<Integer> st = new HashSet<Integer>(); // If 'possible' is true if (possible) { // Printing n numbers // of prime while ((int)st.size() != n) { // checking the form of a+nb num1 = a + (c1 * b); if (isPrime(num1)) { st.add(num1); } c1++; // Checking the form of b+na num2 = b + (c2 * a); if (isPrime(num2)) { st.add(num2); } c2++; } for (int i : st) System.out.print(i + " "); } // If 'possible' is false // return -1 else System.out.print("-1"); } // Driver Code public static void main(String[] args) { int a = 3; int b = 5; int n = 4; findNumbers(a, b, n); } } // This code is contributed by Rajput-Ji |
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Python3
# Python3 implementation of the above approach from math import gcd, sqrt # Utility function to check # whether two numbers is # co-prime or not def coprime(a, b) : if (gcd(a, b) == 1) : return True; else : return False; # Utility function to check # whether a number is prime # or not def isPrime(n) : # Corner case if (n <= 1) : return False; if (n == 2 or n == 3) : return True; # Check from 2 to sqrt(n) for i in range(2, int(sqrt(n)) + 1) : if (n % i == 0) : return False; return True; # finding the Prime numbers def findNumbers(a, b, n) : possible = True; # Checking whether given # numbers are co-prime # or not if (not coprime(a, b)) : possible = False; c1 = 1; c2 = 1; num1 = 0; num2 = 0; # To store the N primes st = set(); # If 'possible' is true if (possible) : # Printing n numbers # of prime while (len(st) != n) : # checking the form of a+nb num1 = a + (c1 * b); if (isPrime(num1)): st.add(num1); c1 += 1; # Checking the form of b+na num2 = b + (c2 * a); if (isPrime(num2)): st.add(num2); c2 += 1; for i in st : print(i, end = " "); # If 'possible' is false # return -1 else : print("-1"); # Driver Code if __name__ == "__main__" : a = 3; b = 5; n = 4; findNumbers(a, b, n); # This code is contributed by AnkitRai01 |
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C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG { static int __gcd(int a, int b) { if (b == 0) return a; return __gcd(b, a % b); } // Utility function to check // whether two numbers is // co-prime or not static bool coprime(int a, int b) { if (__gcd(a, b) == 1) return true; else return false; } // Utility function to check // whether a number is prime // or not static bool isPrime(int n) { // Corner case if (n <= 1) return false; if (n == 2 || n == 3) return true; // Check from 2 to sqrt(n) for (int i = 2; i * i <= n; i++) if (n % i == 0) return false; return true; } // finding the Prime numbers static void findNumbers(int a, int b, int n) { bool possible = true; // Checking whether given // numbers are co-prime // or not if (!coprime(a, b)) possible = false; int c1 = 1; int c2 = 1; int num1, num2; // To store the N primes HashSet<int> st = new HashSet<int>(); // If 'possible' is true if (possible) { // Printing n numbers // of prime while (st.Count != n) { // checking the form of a+nb num1 = a + (c1 * b); if (isPrime(num1)) { st.Add(num1); } c1++; // Checking the form of b+na num2 = b + (c2 * a); if (isPrime(num2)) { st.Add(num2); } c2++; } foreach (int i in st) Console.Write(i + " "); } // If 'possible' is false // return -1 else Console.Write("-1"); } // Driver Code public static void Main(String[] args) { int a = 3; int b = 5; int n = 4; findNumbers(a, b, n); } } // This code is contributed by 29AjayKumar |
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Output:
11 13 17 23
