Check if there exists a number with X factors out of which exactly K are prime
Given two integers X and K, the task is to determine whether there exists a number that has exactly X factors out of which K is prime.
Examples:
Input: X = 8, K = 1
Output: Yes
Explanation:
The number is 128
Factors of 128 = {1, 2, 4, 8, 16, 32, 64, 128} which are 8 in count = X
Among these, only 2 is prime. Therefore count of prime factor = 1 = KInput: X = 4, K = 2
Output: Yes
Explanation:
The number is 6
Factors of 6 = {1, 2, 3, 6} which are 4 in count = X
Among these, only 2 and 3 are prime. Therefore count of prime factor = 2 = K
Approach:
- Suppose a number N has X factors out of which K are prime, say
- Thus, number can be written as
where, the total number of factors is calculated by 
- It is observed that X is a product of “power+1” of the prime factors of the number. Thus, if we are able to divide X into a product of K numbers, then we can form a number with exactly X factors out of which K is prime.
Below is the implementation of the above approach:
C++
// C++ program to check if there exists // a number with X factors // out of which exactly K are prime #include <bits/stdc++.h> using namespace std; // Function to check if such number exists bool check(int X, int K) { int prime, temp, sqr, i; // To store the sum of powers // of prime factors of X which // determines the maximum count // of numbers whose product can form X prime = 0; temp = X; sqr = sqrt(X); // Determining the prime factors of X for (i = 2; i <= sqr; i++) { while (temp % i == 0) { temp = temp / i; prime++; } } // To check if the number is prime if (temp > 2) prime++; // If X is 1, then we cannot form // a number with 1 factor and K // prime factor (as K is atleast 1) if (X == 1) return false; // If X itself is prime then it // can be represented as a power // of only 1 prime factor which // is X itself so we return true if (prime == 1 && K == 1) return true; // If sum of the powers of prime factors // of X is greater than or equal to K, // which means X can be represented as a // product of K numbers, we return true else if (prime >= K) return true; // In any other case, we return false // as we cannot form a number with X // factors and K prime factors else return false; } // Driver code int main() { int X, K; X = 4; K = 2; if (check(X, K)) cout << "Yes"; else cout << "No"; } |
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Java
// Java program to check if there exists // a number with X factors // out of which exactly K are prime import java.util.*; class GFG{ // Function to check if such number exists static boolean check(int X, int K) { int prime, temp, sqr, i; // To store the sum of powers // of prime factors of X which // determines the maximum count // of numbers whose product can form X prime = 0; temp = X; sqr = (int) Math.sqrt(X); // Determining the prime factors of X for (i = 2; i <= sqr; i++) { while (temp % i == 0) { temp = temp / i; prime++; } } // To check if the number is prime if (temp > 2) prime++; // If X is 1, then we cannot form // a number with 1 factor and K // prime factor (as K is atleast 1) if (X == 1) return false; // If X itself is prime then it // can be represented as a power // of only 1 prime factor which // is X itself so we return true if (prime == 1 && K == 1) return true; // If sum of the powers of prime factors // of X is greater than or equal to K, // which means X can be represented as a // product of K numbers, we return true else if (prime >= K) return true; // In any other case, we return false // as we cannot form a number with X // factors and K prime factors else return false; } // Driver code public static void main(String[] args) { int X, K; X = 4; K = 2; if (check(X, K)) System.out.print("Yes"); else System.out.print("No"); } } // This code contributed by Rajput-Ji |
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Python3
# Python3 program to check if there exists # a number with X factors # out of which exactly K are prime from math import sqrt # Function to check if such number exists def check(X,K): # To store the sum of powers # of prime factors of X which # determines the maximum count # of numbers whose product can form X prime = 0 temp = X sqr = int(sqrt(X)) # Determining the prime factors of X for i in range(2,sqr+1,1): while (temp % i == 0): temp = temp // i prime += 1 # To check if the number is prime if (temp > 2): prime += 1 # If X is 1, then we cannot form # a number with 1 factor and K # prime factor (as K is atleast 1) if (X == 1): return False # If X itself is prime then it # can be represented as a power # of only 1 prime factor w0hich # is X itself so we return true if (prime == 1 and K == 1): return True # If sum of the powers of prime factors # of X is greater than or equal to K, # which means X can be represented as a # product of K numbers, we return true elif(prime >= K): return True # In any other case, we return false # as we cannot form a number with X # factors and K prime factors else: return False # Driver code if __name__ == '__main__': X = 4 K = 2 if (check(X, K)): print("Yes") else: print("No") # This code is contributed by Surendra_Gangwar |
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C#
// C# program to check if there exists // a number with X factors // out of which exactly K are prime using System; class GFG{ // Function to check if such number exists static bool check(int X, int K) { int prime, temp, sqr, i; // To store the sum of powers // of prime factors of X which // determines the maximum count // of numbers whose product can form X prime = 0; temp = X; sqr = Convert.ToInt32(Math.Sqrt(X)); // Determining the prime factors of X for (i = 2; i <= sqr; i++) { while (temp % i == 0) { temp = temp / i; prime++; } } // To check if the number is prime if (temp > 2) prime++; // If X is 1, then we cannot form // a number with 1 factor and K // prime factor (as K is atleast 1) if (X == 1) return false; // If X itself is prime then it // can be represented as a power // of only 1 prime factor which // is X itself so we return true if (prime == 1 && K == 1) return true; // If sum of the powers of prime factors // of X is greater than or equal to K, // which means X can be represented as a // product of K numbers, we return true else if (prime >= K) return true; // In any other case, we return false // as we cannot form a number with X // factors and K prime factors else return false; } // Driver code static public void Main () { int X, K; X = 4; K = 2; if (check(X, K)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } // This code is contributed by shubhamsingh10 |
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Output:
Yes
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