Check if a number has prime count of divisors
Last Updated :
12 Jul, 2025
Given an integer N, the task is to check if the count of divisors of N is prime or not.
Examples:
Input: N = 13
Output: Yes
The divisor count is 2 (1 and 13) which is prime.
Input: N = 8
Output: No
The divisors are 1, 2, 4 and 8.
Approach: Please read this article to find the count of divisors of a number. So find the maximum value of i for every prime divisor p such that N % (pi) = 0. So the count of divisors gets multiplied by (i + 1). The count of divisors will be (i1 + 1) * (i2 + 1) * ... * (ik + 1).
It can now be seen that there can only be one prime divisor for the maximum i and if N % pi = 0 then (i + 1) should be prime. The primality can be checked in sqrt(n) time and the prime factors can also be found in sqrt(n) time. So the overall time complexity will be O(sqrt(n)).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true
// if n is prime
bool Prime(int n)
{
// There is no prime
// less than 2
if (n < 2)
return false;
// Run a loop from 2 to sqrt(n)
for (int i = 2; i <= sqrt(n); i++)
// If there is any factor
if (n % i == 0)
return false;
return true;
}
// Function that returns true if n
// has a prime count of divisors
bool primeCountDivisors(int n)
{
if (n < 2)
return false;
// Find the prime factors
for (int i = 2; i <= sqrt(n); i++)
if (n % i == 0) {
// Find the maximum value of i for every
// prime divisor p such that n % (p^i) == 0
long a = n, c = 0;
while (a % i == 0) {
a /= i;
c++;
}
// If c+1 is a prime number and a = 1
if (a == 1 && Prime(c + 1))
return true;
// The number cannot have two factors
// to have count of divisors prime
else
return false;
}
// Else the number is prime so
// it has only two divisors
return true;
}
// Driver code
int main()
{
int n = 13;
if (primeCountDivisors(n))
cout << "Yes";
else
cout << "No";
return 0;
}
// Java implementation of the approach
class GFG
{
// Function that returns true
// if n is prime
static boolean Prime(int n)
{
// There is no prime
// less than 2
if (n < 2)
return false;
// Run a loop from 2 to sqrt(n)
for (int i = 2; i <= (int)Math.sqrt(n); i++)
// If there is any factor
if (n % i == 0)
return false;
return true;
}
// Function that returns true if n
// has a prime count of divisors
static boolean primeCountDivisors(int n)
{
if (n < 2)
return false;
// Find the prime factors
for (int i = 2; i <= (int)Math.sqrt(n); i++)
if (n % i == 0)
{
// Find the maximum value of i for every
// prime divisor p such that n % (p^i) == 0
long a = n, c = 0;
while (a % i == 0)
{
a /= i;
c++;
}
// If c+1 is a prime number and a = 1
if (a == 1 && Prime((int)c + 1) == true)
return true;
// The number cannot have two factors
// to have count of divisors prime
else
return false;
}
// Else the number is prime so
// it has only two divisors
return true;
}
// Driver code
public static void main (String[] args)
{
int n = 13;
if (primeCountDivisors(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by AnkitRai01
# Python3 implementation of the approach
from math import sqrt
# Function that returns true
# if n is prime
def Prime(n) :
# There is no prime
# less than 2
if (n < 2) :
return False;
# Run a loop from 2 to sqrt(n)
for i in range(2, int(sqrt(n)) + 1) :
# If there is any factor
if (n % i == 0) :
return False;
return True;
# Function that returns true if n
# has a prime count of divisors
def primeCountDivisors(n) :
if (n < 2) :
return False;
# Find the prime factors
for i in range(2, int(sqrt(n)) + 1) :
if (n % i == 0) :
# Find the maximum value of i for every
# prime divisor p such that n % (p^i) == 0
a = n; c = 0;
while (a % i == 0) :
a //= i;
c += 1;
# If c + 1 is a prime number and a = 1
if (a == 1 and Prime(c + 1)) :
return True;
# The number cannot have two factors
# to have count of divisors prime
else :
return False;
# Else the number is prime so
# it has only two divisors
return True;
# Driver code
if __name__ == "__main__" :
n = 13;
if (primeCountDivisors(n)) :
print("Yes");
else :
print("No");
# This code is contributed by AnkitRai01
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true
// if n is prime
static bool Prime(int n)
{
// There is no prime
// less than 2
if (n < 2)
return false;
// Run a loop from 2 to sqrt(n)
for (int i = 2; i <= (int)Math.Sqrt(n); i++)
// If there is any factor
if (n % i == 0)
return false;
return true;
}
// Function that returns true if n
// has a prime count of divisors
static bool primeCountDivisors(int n)
{
if (n < 2)
return false;
// Find the prime factors
for (int i = 2; i <= (int)Math.Sqrt(n); i++)
if (n % i == 0)
{
// Find the maximum value of i for every
// prime divisor p such that n % (p^i) == 0
long a = n, c = 0;
while (a % i == 0)
{
a /= i;
c++;
}
// If c+1 is a prime number and a = 1
if (a == 1 && Prime((int)c + 1) == true)
return true;
// The number cannot have two factors
// to have count of divisors prime
else
return false;
}
// Else the number is prime so
// it has only two divisors
return true;
}
// Driver code
public static void Main()
{
int n = 13;
if (primeCountDivisors(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by AnkitRai01
<script>
// Javascript implementation of the approach
// Function that returns true
// if n is prime
function Prime(n)
{
// There is no prime
// less than 2
if (n < 2)
return false;
// Run a loop from 2 to sqrt(n)
for(var i = 2; i <= Math.sqrt(n); i++)
// If there is any factor
if (n % i == 0)
return false;
return true;
}
// Function that returns true if n
// has a prime count of divisors
function primeCountDivisors( n)
{
if (n < 2)
return false;
// Find the prime factors
for(var i = 2; i <= Math.sqrt(n); i++)
if (n % i == 0)
{
// Find the maximum value of i for every
// prime divisor p such that n % (p^i) == 0
var a = n, c = 0;
while (a % i == 0)
{
a /= i;
c++;
}
// If c+1 is a prime number and a = 1
if (a == 1 && Prime(c + 1))
return true;
// The number cannot have two factors
// to have count of divisors prime
else
return false;
}
// Else the number is prime so
// it has only two divisors
return true;
}
// Driver rcode
n = 13;
if (primeCountDivisors(n))
document.write("Yes");
else
document.write("No");
// This code is contributed by SoumikMondal
</script>
Time Complexity: O(sqrt(n)), as we are using a loop to traverse sqrt (n) times. Where n is the integer given as input.
Auxiliary Space: O(1), as we are not using any extra space.
Approach 2: HashMap:
- The approach uses a map to count the frequency of each prime factor. A map is a data structure in C++ that stores key-value pairs. In this case, the key is the prime factor, and the value is the frequency of that prime factor. The map is used because it allows us to keep track of the frequency of each prime factor efficiently without needing to know the prime factors beforehand.
- The approach then uses a for loop to iterate over all the prime factors and count their frequency using the map. It also checks if there is any remaining factor greater than 1 after dividing by all the prime factors up to the square root of the number. If there is such a factor, it is also included in the map with a frequency of 1.
- After counting the frequency of each prime factor, the approach uses another for loop to calculate the total number of divisors of the number using the formula (f1+1) * (f2+1) * ... * (fn+1), where f1, f2, ..., fn are the frequencies of each prime factor in the factorization of the number. This formula works because for each prime factor, we have (frequency + 1) choices for how many times to include that prime factor in a divisor. We multiply all these choices together to get the total number of divisors.
- Finally, the approach checks if the total number of divisors is a prime number using another for loop that checks if the number is divisible by any number from 2 up to the square root of the number of divisors. If it is divisible by any of these numbers, then it is not a prime number and the function returns false. If none of these numbers divide the number of divisors, then it is a prime number, and the function returns true.
Here is the code of this approach:
C++
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if n
// has a prime count of divisors
bool primeCountDivisors(int n)
{
if (n < 2)
return false;
// Find all the prime factors of n
vector<int> primes;
for (int i = 2; i <= sqrt(n); i++) {
while (n % i == 0) {
primes.push_back(i);
n /= i;
}
}
if (n > 1) {
primes.push_back(n);
}
// Count the frequency of each prime factor
map<int, int> freq;
for (int prime : primes) {
freq[prime]++;
}
// Calculate the number of divisors of n
int numDivisors = 1;
for (auto it : freq) {
numDivisors *= (it.second + 1);
}
// Check if the number of divisors is a prime number or not
if (numDivisors < 2)
return false;
for (int i = 2; i <= sqrt(numDivisors); i++) {
if (numDivisors % i == 0)
return false;
}
return true;
}
// Driver code
int main()
{
int n = 13;
if (primeCountDivisors(n))
cout << "Yes";
else
cout << "No";
return 0;
}
import java.util.HashMap;
import java.util.Map;
public class PrimeCountDivisors {
public static boolean primeCountDivisors(int n) {
if (n < 2) {
return false;
}
// Find all the prime factors of n and their occurrences using a HashMap
Map<Integer, Integer> primes = new HashMap<>();
for (int i = 2; i <= Math.sqrt(n); i++) {
while (n % i == 0) {
// Increment the count of the prime factor in the HashMap
primes.put(i, primes.getOrDefault(i, 0) + 1);
n /= i; // Reduce n by dividing it by the prime factor
}
}
// If n is still greater than 1, it is a prime factor itself
if (n > 1) {
primes.put(n, 1);
}
// Calculate the number of divisors of n using the prime factorization
int numDivisors = 1;
for (Map.Entry<Integer, Integer> entry : primes.entrySet()) {
numDivisors *= (entry.getValue() + 1);
}
// Check if the number of divisors is a prime number or not
// If the number of divisors is less than 2, it is not prime
if (numDivisors < 2) {
return false;
}
// Check if the number of divisors is prime by checking for divisors from 2 to sqrt(numDivisors)
for (int i = 2; i <= Math.sqrt(numDivisors); i++) {
if (numDivisors % i == 0) {
return false; // If a divisor is found, numDivisors is not a prime number
}
}
// If no divisors are found other than 1 and numDivisors, it is a prime number
return true;
}
public static void main(String[] args) {
int n = 13;
// Check if the number of divisors of 'n' is a prime number
if (primeCountDivisors(n)) {
System.out.println("Yes"); // If it's prime, print "Yes"
} else {
System.out.println("No"); // If it's not prime, print "No"
}
}
}
import math
# Function that returns True if n has a prime count of divisors
def primeCountDivisors(n):
if n < 2:
return False
# Find all the prime factors of n
primes = []
i = 2
while i <= math.sqrt(n):
while n % i == 0:
primes.append(i)
n //= i
i += 1
if n > 1:
primes.append(n)
# Count the frequency of each prime factor
freq = {}
for prime in primes:
freq[prime] = freq.get(prime, 0) + 1
# Calculate the number of divisors of n
numDivisors = 1
for count in freq.values():
numDivisors *= (count + 1)
# Check if the number of divisors is a prime number or not
if numDivisors < 2:
return False
for i in range(2, int(math.sqrt(numDivisors)) + 1):
if numDivisors % i == 0:
return False
return True
# Driver code
if __name__ == "__main__":
n = 13
if primeCountDivisors(n):
print("Yes")
else:
print("No")
using System;
using System.Collections.Generic;
class Program {
static bool PrimeCountDivisors(int n)
{
if (n < 2)
return false;
// Find all the prime factors of n
List<int> primes = new List<int>();
for (int i = 2; i <= Math.Sqrt(n); i++) {
while (n % i == 0) {
primes.Add(i);
n /= i;
}
}
if (n > 1) {
primes.Add(n);
}
// Count the frequency of each prime factor
Dictionary<int, int> freq
= new Dictionary<int, int>();
foreach(int prime in primes)
{
if (freq.ContainsKey(prime)) {
freq[prime]++;
}
else {
freq[prime] = 1;
}
}
// Calculate the number of divisors of n
int numDivisors = 1;
foreach(KeyValuePair<int, int> pair in freq)
{
numDivisors *= (pair.Value + 1);
}
// Check if the number of divisors is a prime number
// or not
if (numDivisors < 2)
return false;
for (int i = 2; i <= Math.Sqrt(numDivisors); i++) {
if (numDivisors % i == 0)
return false;
}
return true;
}
static void Main()
{
int n = 13;
if (PrimeCountDivisors(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by sarojmcy2e
// Function that returns true if n
// has a prime count of divisors
function primeCountDivisors(n) {
if (n < 2) return false;
// Find all the prime factors of n
let primes = [];
for (let i = 2; i <= Math.sqrt(n); i++) {
while (n % i === 0) {
primes.push(i);
n /= i;
}
}
if (n > 1) {
primes.push(n);
}
// Count the frequency of each prime factor
let freq = {};
for (let prime of primes) {
freq[prime] = (freq[prime] || 0) + 1;
}
// Calculate the number of divisors of n
let numDivisors = 1;
for (let key in freq) {
numDivisors *= (freq[key] + 1);
}
// Check if the number of divisors is a prime number or not
if (numDivisors < 2) return false;
for (let i = 2; i <= Math.sqrt(numDivisors); i++) {
if (numDivisors % i === 0) return false;
}
return true;
}
// Driver code
let n = 13;
if (primeCountDivisors(n)) {
console.log("Yes");
} else {
console.log("No");
}
Output:
Yes
Time Complexity: O(sqrt(N)), as we are using a loop to traverse sqrt (N) times. Where n is the integer given as input.
Auxiliary Space: O(LogN), as we are not using any extra space.
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