# Expressing factorial n as sum of consecutive numbers

Given two numbers N and M. Find the number of ways in which factorial N can be expressed as a sum of two or more consecutive numbers. Print the result modulo M.

Examples:

Input : N = 3, M = 7
Output : 1
Explanation:  3! can be expressed
in one way, i.e. 1 + 2 + 3 = 6.
Hence 1 % 7 = 1

Input : N = 4, M = 7
Output : 1
Explanation:  4! can be expressed
in one way, i.e. 7 + 8 + 9 = 24
Hence 1 % 7 = 1


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

A simple solution is to first compute factorial, then count number of ways to represent factorial as sum of consecutive numbers using Count ways to express a number as sum of consecutive numbers. This solution causes overflow.

Below is a better solution to avoid overflow.
Let us consider that sum of r consecutive numbers be expressed as:
(a + 1) + (a + 2) + (a + 3) + … + (a + r), which simplifies as (r * (r + 2*a + 1)) / 2
Hence, (a + 1) + (a + 2) + (a + 3) + … + (a + r) = (r * (r + 2*a + 1)) / 2. Since the above expression is equal to factorial N, we write it as
2 * N! = r * (r + 2*a + 1)
Instead of counting all the pairs (r, a), we will count all pairs (r, r + 2*a + 1). Now, we are just counting all ordered pairs (X, Y) with XY = 2 * N! where X < Y and X, Y have different parity, that means if (r) is even, (r + 2*a + 1) is odd or if (r) is odd then (r + 2*a + 1) is even. This is equivalent to finding the odd divisors of 2 * N! which will be same as odd divisors of N!.
For counting the number of divisors in N!, we calculate the power of primes in factorization and total count of divisors become (p1 + 1) * (p2 + 1) * … * (pn + 1). To calculate the largest power of a prime in N!, we will use legendre’s formula. Below is the implementation of the above approach.

## C++

 // CPP program to count number of  // ways we can express a factorial  // as sum of consecutive numbers  #include  using namespace std;     #define MAX 50002     vector<int> primes;     // sieve of Eratosthenes to compute  // the prime numbers  void sieve()  {      bool isPrime[MAX];      memset(isPrime, true, sizeof(isPrime));         for (int p = 2; p * p < MAX; p++) {          if (isPrime[p] == true) {              for (int i = p * 2; i < MAX; i += p)                  isPrime[i] = false;          }      }         // Store all prime numbers      for (int p = 2; p < MAX; p++)          if (isPrime[p])              primes.push_back(p);  }     // function to calculate the largest  // power of a prime in a number  long long int power(long long int x,                      long long int y)  {      long long int count = 0;      long long int z = y;      while (x >= z) {          count += (x / z);          z *= y;      }      return count;  }     // Modular multiplication to avoid  // the overflow of multiplication  // Please see below for details  // https://www.geeksforgeeks.org/how-to-avoid-overflow-in-modular-multiplication/  long long int modMult(long long int a,                        long long int b,                        long long int mod)  {      long long int res = 0;      a = a % mod;      while (b > 0) {          if (b % 2 == 1)              res = (res + a) % mod;          a = (a * 2) % mod;          b /= 2;      }      return res % mod;  }     // Returns count of ways to express n!  // as sum of consecutives.  long long int countWays(long long int n,                          long long int m)  {      long long int ans = 1;         // We skip 2 (First prime) as we need to      // consider only odd primes      for (int i = 1; i < primes.size(); i++) {             // compute the largest power of prime          long long int powers = power(n, primes[i]);             // if the power of current prime number          // is zero in N!, power of primes greater          // than current prime number will also          // be zero, so break out from the loop          if (powers == 0)              break;             // multiply the result at every step          ans = modMult(ans, powers + 1, m) % m;      }         // subtract 1 to exclude the case of 1      // being an odd divisor      if (((ans - 1) % m) < 0)          return (ans - 1 + m) % m;      else         return (ans - 1) % m;  }     // Driver Code  int main()  {      sieve();      long long int n = 4, m = 7;      cout << countWays(n, m);      return 0;  }

## Java

 // Java program to count number of  // ways we can express a factorial  // as sum of consecutive numbers  import java.util.*;     class GFG {             static int MAX = 50002;      static ArrayList primes                    = new ArrayList();                          // sieve of Eratosthenes to compute      // the prime numbers      public static void sieve()      {                     boolean isPrime[] = new boolean[MAX];                     for(int i = 0; i < MAX; i++)              isPrime[i] = true;                         for (int p = 2; p * p < MAX; p++) {              if (isPrime[p] == true) {                  for (int i = p * 2; i < MAX; i += p)                      isPrime[i] = false;              }          }                     // Store all prime numbers          for (int p = 2; p < MAX; p++)              if (isPrime[p] == true)                  primes.add(p);      }                 // function to calculate the largest      // power of a prime in a number      public static int power(int x, int y)      {          int count = 0;          int z = y;          while (x >= z) {              count += (x / z);              z *= y;          }                     return count;      }             // Modular multiplication to avoid      // the overflow of multiplication      // Please see below for details      // https://www.geeksforgeeks.org/how-to-avoid-overflow-in-modular-multiplication/      public static int modMult(int a, int b, int mod)      {          int res = 0;          a = a % mod;          while (b > 0) {              if (b % 2 == 1)                  res = (res + a) % mod;              a = (a * 2) % mod;              b /= 2;          }                     return res % mod;      }             // Returns count of ways to express n!      // as sum of consecutives.      public static int countWays(int n, int m)      {           int ans = 1;                 // We skip 2 (First prime) as we need to          // consider only odd primes          for (int i = 1; i < primes.size(); i++) {                     // compute the largest power of prime              int powers = power(n, primes.get(i));                     // if the power of current prime number              // is zero in N!, power of primes greater              // than current prime number will also              // be zero, so break out from the loop              if (powers == 0)                  break;                     // multiply the result at every step              ans = modMult(ans, powers + 1, m) % m;          }                 // subtract 1 to exclude the case of 1          // being an odd divisor          if (((ans - 1) % m) < 0)              return (ans - 1 + m) % m;          else             return (ans - 1) % m;      }             //Driver function      public static void main (String[] args) {                     sieve();                     int n = 4, m = 7;                      System.out.println(countWays(n,m));      }  }     // This code is contributed by akash1295.

## Python 3

 # Python 3 program to count number of  # ways we can express a factorial  # as sum of consecutive numbers  MAX = 50002;     primes = []     # sieve of Eratosthenes to compute  # the prime numbers  def sieve():      isPrime = [True]*(MAX)             p = 2     while p * p < MAX :          if (isPrime[p] == True):              for i in range( p * 2,MAX, p):                  isPrime[i] = False         p+=1        # Store all prime numbers      for p in range( 2,MAX):          if (isPrime[p]):              primes.append(p)     # function to calculate the largest  # power of a prime in a number  def power( x, y):         count = 0     z = y      while (x >= z):          count += (x // z)          z *= y             return count     # Modular multiplication to avoid  # the overflow of multiplication  # Please see below for details  # https://www.geeksforgeeks.org/how-to-avoid-overflow-in-modular-multiplication/  def modMult(a, b,mod):      res = 0     a = a % mod      while (b > 0):          if (b % 2 == 1):              res = (res + a) % mod          a = (a * 2) % mod          b //= 2            return res % mod     # Returns count of ways to express n!  # as sum of consecutives.  def countWays(n,m):      ans = 1        # We skip 2 (First prime) as we need to      # consider only odd primes      for i in range(1,len(primes)):             # compute the largest power of prime          powers = power(n, primes[i])             # if the power of current prime number          # is zero in N!, power of primes greater          # than current prime number will also          # be zero, so break out from the loop          if (powers == 0):              break            # multiply the result at every step          ans = modMult(ans, powers + 1, m) % m                # subtract 1 to exclude the case of 1      # being an odd divisor      if (((ans - 1) % m) < 0):          return (ans - 1 + m) % m      else:          return (ans - 1) % m     # Driver Code  if __name__ == "__main__":      sieve()      n = 4     m = 7     print(countWays(n, m))         # This code is contributed by ChitraNayal

## C#

 // C# program to count number of   // ways we can express a factorial   // as sum of consecutive numbers   using System ;  using System.Collections;     class GFG {              static int MAX = 50002;       static ArrayList primes = new ArrayList ();                          // sieve of Eratosthenes to compute       // the prime numbers       public static void sieve()       {                      bool []isPrime = new bool[MAX];                      for(int i = 0; i < MAX; i++)               isPrime[i] = true;                          for (int p = 2; p * p < MAX; p++) {               if (isPrime[p] == true) {                   for (int i = p * 2; i < MAX; i += p)                       isPrime[i] = false;               }           }                      // Store all prime numbers           for (int p = 2; p < MAX; p++)               if (isPrime[p] == true)                   primes.Add(p);       }                  // function to calculate the largest       // power of a prime in a number       public static int power_prime(int x, int y)       {           int count = 0;           int z = y;           while (x >= z) {               count += (x / z);               z *= y;           }                      return count;       }              // Modular multiplication to avoid       // the overflow of multiplication       // Please see below for details       // https://www.geeksforgeeks.org/how-to-avoid-overflow-in-modular-multiplication/       public static int modMult(int a, int b, int mod)       {           int res = 0;           a = a % mod;           while (b > 0) {               if (b % 2 == 1)                   res = (res + a) % mod;               a = (a * 2) % mod;               b /= 2;           }                      return res % mod;       }              // Returns count of ways to express n!       // as sum of consecutives.       public static int countWays(int n, int m)       {           int ans = 1;                  // We skip 2 (First prime) as we need to           // consider only odd primes           for (int i = 1; i < primes.Count; i++) {                      // compute the largest power of prime               int powers = power_prime(n, Convert.ToInt32(primes[i]));                      // if the power of current prime number               // is zero in N!, power of primes greater               // than current prime number will also               // be zero, so break out from the loop               if (powers == 0)                   break;                      // multiply the result at every step               ans = modMult(ans, powers + 1, m) % m;           }                  // subtract 1 to exclude the case of 1           // being an odd divisor           if (((ans - 1) % m) < 0)               return (ans - 1 + m) % m;           else             return (ans - 1) % m;       }              //Driver function       public static void Main () {                      sieve();                      int n = 4, m = 7;                      Console.WriteLine(countWays(n,m));       }   }      // This code is contributed by Ryuga

Output:

1


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