# 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.


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have discussed different approaches for printing all divisors (here and here). Here 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 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:

 // CPP program for finding number of divisor  #include     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 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)

 # 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# 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

 

Output:
8


Reference : Number of divisors.

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