Total number of divisors for a given number (original) (raw)

Last Updated : 15 Oct, 2024

Given a positive integer n, we have to find the total number of divisors for n.

**Examples:

**Input : n = 25
**Output : 3
Divisors are 1, 5 and 25.

**Input : n = 24
**Output : 8
Divisors are 1, 2, 3, 4, 6, 8
12 and 24.

We have discussed different approaches for printing all divisors (here and here). Here the task is simpler, we need to count divisors. First of all store all primes from 2 to max_size in an array so that we should only check for the prime divisors. Now we will only wish to calculate the factorization of n in the following form:
n = \prod_{i=1}^{n} a_{i}^{p_{i}} = a_{1}^{p_{1}}\times a_{2}^{p_{2}}\times a_3^{p_{3}}\times.....\times a_{n}^{p_{n}} where a i are prime factors and p i are integral power of them. So, for this factorization, we have formula to find total number of divisor of n and that is: \prod_{i=1}^{n} (p_{i}+1)= (p_{0}+1)\times (p_{1}+1) \times......(p_{n}+1)

C++ `

// CPP program for finding number of divisor #include <bits/stdc++.h>

using namespace std;

// program for finding no. of divisors int divCount(int n) { // sieve method for prime calculation bool hash[n + 1]; memset(hash, true, sizeof(hash)); for (int p = 2; p * p < n; p++) if (hash[p] == true) for (int i = p * 2; i < n; i += p) hash[i] = false;

// Traversing through all prime numbers
int total = 1;
for (int p = 2; p <= n; p++) {
    if (hash[p]) {

        // calculate number of divisor
        // with formula total div = 
        // (p1+1) * (p2+1) *.....* (pn+1)
        // where n = (a1^p1)*(a2^p2).... 
        // *(an^pn) ai being prime divisor
        // for n and pi are their respective 
        // power in factorization
        int count = 0;
        if (n % p == 0) {
            while (n % p == 0) {
                n = n / p;
                count++;
            }
            total = total * (count + 1);
        }
    }
}
return total;

}

// driver program int main() { int n = 24; cout << divCount(n); return 0; }

Java

// Java program for finding // number of divisor import java.io.; import java.util.; import java.lang.*;

class GFG { // program for finding // no. of divisors static int divCount(int n) { // sieve method for prime calculation boolean hash[] = new boolean[n + 1]; Arrays.fill(hash, true); for (int p = 2; p * p < n; p++) if (hash[p] == true) for (int i = p * 2; i < n; i += p) hash[i] = false;

// Traversing through 
// all prime numbers
int total = 1;
for (int p = 2; p <= n; p++) 
{
    if (hash[p])
    {

        // calculate number of divisor
        // with formula total div = 
        // (p1+1) * (p2+1) *.....* (pn+1)
        // where n = (a1^p1)*(a2^p2).... 
        // *(an^pn) ai being prime divisor
        // for n and pi are their respective 
        // power in factorization
        int count = 0;
        if (n % p == 0) 
        {
            while (n % p == 0) 
            {
                n = n / p;
                count++;
            }
            total = total * (count + 1);
        }
    }
}
return total;

}

// Driver Code public static void main(String[] args) { int n = 24; System.out.print(divCount(n)); } }

// This code is contributed // by Akanksha Rai(Abby_akku)

Python

Python3 program for finding

number of divisor

program for finding

no. of divisors

def divCount(n):

# sieve method for
# prime calculation
hh = [1] * (n + 1);

p = 2;
while((p * p) < n):
    if (hh[p] == 1):
        for i in range((p * 2), n, p):
            hh[i] = 0;
    p += 1;

# Traversing through 
# all prime numbers
total = 1;
for p in range(2, n + 1):
    if (hh[p] == 1):

        # calculate number of divisor
        # with formula total div = 
        # (p1+1) * (p2+1) *.....* (pn+1)
        # where n = (a1^p1)*(a2^p2).... 
        # *(an^pn) ai being prime divisor
        # for n and pi are their respective 
        # power in factorization
        count = 0;
        if (n % p == 0):
            while (n % p == 0):
                n = int(n / p);
                count += 1;
            total *= (count + 1);
            
return total;

Driver Code

n = 24; print(divCount(n));

This code is contributed by mits

C#

// C# program for finding // number of divisor using System;

class GFG { // program for finding // no. of divisors static int divCount(int n) { // sieve method for prime calculation bool[] hash = new bool[n + 1]; for (int p = 2; p * p < n; p++) if (hash[p] == false) for (int i = p * 2; i < n; i += p) hash[i] = true;

// Traversing through 
// all prime numbers
int total = 1;
for (int p = 2; p <= n; p++) 
{
    if (hash[p] == false)
    {

        // calculate number of divisor
        // with formula total div = 
        // (p1+1) * (p2+1) *.....* (pn+1)
        // where n = (a1^p1)*(a2^p2).... 
        // *(an^pn) ai being prime divisor
        // for n and pi are their respective 
        // power in factorization
        int count = 0;
        if (n % p == 0) 
        {
            while (n % p == 0) 
            {
                n = n / p;
                count++;
            }
            total = total * (count + 1);
        }
    }
}
return total;

}

// Driver Code public static void Main() { int n = 24; Console.WriteLine(divCount(n)); } }

// This code is contributed // by mits

JavaScript

PHP

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