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# Total number of divisors for a given number

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 = = where ai are prime factors and pi are integral power of them.
So, for this factorization we have formula to find total number of divisor of n and that is: ## C++

 // CPP program for finding number of divisor#include   using namespace std;  // program for finding no. of divisorsint 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 programint main(){    int n = 24;    cout << divCount(n);    return 0;}

## Java

 // Java program for finding// number of divisorimport java.io.*;import java.util.*;import java.lang.*;  class GFG{// program for finding // no. of divisorsstatic 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 Codepublic static void main(String[] args){    int n = 24;    System.out.print(divCount(n));}}  // This code is contributed // by Akanksha Rai(Abby_akku)

## Python3

 # Python3 program for finding # number of divisor  # program for finding # no. of divisorsdef divCount(n):      # sieve method for    # prime calculation    hh =  * (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 Coden = 24;print(divCount(n));  # This code is contributed by mits

## C#

 // C# program for finding// number of divisorusing System;  class GFG{// program for finding // no. of divisorsstatic 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 Codepublic static void Main(){    int n = 24;    Console.WriteLine(divCount(n));}}  // This code is contributed // by mits

## PHP

 

## Javascript

 

Output:

8

Reference : Number of divisors.