Given a range [L, R], and a prime number P. We are required to find the sum of the highest power of P in all numbers from L to R.

Examples:

Input : L = 1, R = 10, P = 2 Output : 8 There are 10 integers in the range, and: In 1 the highest power of 2 is 0 In 2 the highest power of 2 is 1 In 3 the highest power of 2 is 0 Similarly the highest powers of 2 in 4, 5, 6, 7, 8, 9, 10 are 2, 0, 1, 0, 3, 0, 1 respectively. Input : L = 10, R = 20, P = 7 Output : 1 There are 11 integers in the range, and: In 10 the highest power of 7 is 0 In 11 the highest power of 7 is 0 In 12 the highest power of 7 is 0 Similarly the highest power of 7 in 13, 14, 15, 16, 17, 18, 19, 20 and 10 is 0, 1, 0, 0, 0, 0, 0 and 0 respectively.

**Native Approach : **Run a loop from L to R. For every number, check if it is divisible by P. If yes, then find the largest power of P that divides current number. If yes, then add this value to result.

## C++

// Simple CPP Program to find sum of largest // divisible powers of P in [L, R] #include <bits/stdc++.h> using namespace std; int sumOfDivisblePowers(int L, int R, int P) { // Traverse through all numbers from L to R int res = 0; for (int i = L; i <= R; i++) { // If P divides current number x, add // largest power of P that divides x. int x = i; while (x % P == 0) { res++; x /= P; } } return res; } // Driver code int main() { int L = 1, R = 10, P = 2; cout << sumOfDivisblePowers(L, R, P); return 0; }

## Java

// Simple Java Program to find sum of largest // divisible powers of P in [L, R] class GFG { // Utility Function static int sumOfDivisblePowers(int L, int R, int P) { // Traverse through all numbers from L to R int res = 0; for (int i = L; i <= R; i++) { // If P divides current number x, add // largest power of P that divides x. int x = i; while (x % P == 0) { res++; x /= P; } } return res; } // Driver code public static void main (String[] args) { int L = 1, R = 10, P = 2; System.out.println(sumOfDivisblePowers(L, R, P)); } } // This code is contributed by Anant Agarwal.

## Python3

# Simple Python 3 Program to find sum of largest # divisible powers of P in [L, R] def sumOfDivisblePowers(L, R, P): # Traverse through all numbers from L to R res = 0 for i in range(L, R + 1): # If P divides current number x, add # largest power of P that divides x. x = i while (x % P == 0): res += 1 x /= P return res # Driver code L = 1 R = 10 P = 2 print(sumOfDivisblePowers(L, R, P)) # This code is contributed by Smitha Dinesh Semwal

## C#

// Simple C# Program to find sum of // largest divisible powers of P in // [L, R] using System; class GFG { // Utility Function static int sumOfDivisblePowers(int L, int R, int P) { // Traverse through all numbers // from L to R int res = 0; for (int i = L; i <= R; i++) { // If P divides current number, // x, add largest power of P // that divides x. int x = i; while (x % P == 0) { res++; x /= P; } } return res; } // Driver code public static void Main () { int L = 1, R = 10, P = 2; Console.WriteLine(sumOfDivisblePowers(L, R, P)); } } // This code is contributed by vt_m.

## PHP

<?php // Simple PHP Program to // find sum of largest // divisible powers of // P in [L, R] // function return result function sumOfDivisblePowers($L, $R, $P) { // Traverse through all // numbers from L to R $res = 0; for ($i = $L; $i <= $R; $i++) { // If P divides current // number x, add // largest power of // P that divides x. $x = $i; while ($x % $P == 0) { $res++; $x /= $P; } } return $res; } // Driver Code $L = 1; $R = 10; $P = 2; echo sumOfDivisblePowers($L, $R, $P); // This code is contributed by ajit ?>

**Output:**

8

**Efficient approach:** The idea is to use to Legendre’s formula.

The largest possible power of P that divides factorial of a number x is ⌊x/p⌋ + ⌊x/(p

^{2})⌋ + ⌊x/(p^{3})⌋ + ……Using above, we can find the largest power of P that divides R! (or 1 * 2 * 3…. R).

Sum of largest powers from L to R = Largest Power of P that divides R! (or 1 * 2 … L-1 * L * … R) – Largest Power of P that divides (L-1)! (or 1 * 2 … L-1)

## C++

// Efficient CPP Program to find sum of // largest divisible powers of P in [L, R] #include <bits/stdc++.h> using namespace std; // Implements Lagrange's theorem (Finds largest // power of P that divides x! int largestPower(int x, int P) { // Calculate x/p + x/(p^2) + x/(p^3) + .... int res = 0; while (x) { x /= P; res += x; } return res; } // Returns sum of largest powers of P that divide // numbers in range from L to R. int sumOfDivisblePowers(int L, int R, int P) { return largestPower(R, P) - largestPower(L-1, P); } // Driver code int main() { int L = 1, R = 10, P = 2; cout << sumOfDivisblePowers(L, R, P); return 0; }

## Java

// Efficient CPP Program to find sum of // largest divisible powers of P in [L, R] class GFG { // Implements Lagrange's theorem // (Finds largest power of P that // divides x! static int largestPower(int x, int P) { // Calculate x/p + x/(p^2) + x/(p^3) + .... int res = 0; while(x != 0) { x /= P; res += x; } return res; } // Returns sum of largest powers // of P that divide numbers in // range from L to R. static int sumOfDivisblePowers(int L, int R, int P) { return largestPower(R, P) - largestPower(L-1, P); } // Driver code public static void main (String[] args) { int L = 1, R = 10, P = 2; System.out.println(sumOfDivisblePowers(L, R, P)); } } // This code is contributed by Anant Agarwal.

## Python3

# Efficient Python 3 Program to find sum of # largest divisible powers of P in [L, R] # Implements Lagrange's theorem (Finds largest # power of P that divides x! def largestPower(x, P): # Calculate x/p + x/(p^2) + x/(p^3) + .... res = 0 while (x): x = int(x / P) res += x return res # Returns sum of largest powers of P that divide # numbers in range from L to R. def sumOfDivisblePowers( L, R, P) : return largestPower(R, P) - largestPower(L-1, P) # Driver code L = 1 R = 10 P = 2 print(sumOfDivisblePowers(L, R, P)) # This code is contributed by Smitha Dinesh Semwal

## C#

// Efficient C# Program to find sum of // largest divisible powers of P in [L, R] using System; class GFG { // Implements Lagrange's theorem (Finds // largest power of P that divides x! static int largestPower(int x, int P) { // Calculate x/p + x/(p^2) + // x/(p^3) + .... int res = 0; while (x>0) { x /= P; res += x; } return res; } // Returns sum of largest powers of // P that divide numbers in range // from L to R. static int sumOfDivisblePowers(int L, int R, int P) { return largestPower(R, P) - largestPower(L-1, P); } // Driver Program to test above // function static void Main() { int L = 1, R = 10, P = 2; Console.Write( sumOfDivisblePowers(L, R, P)); } } // This code is contributed by Anuj_67

## PHP

<?php // Efficient php Program to find sum of // largest divisible powers of P in [L, R] // Implements Lagrange's theorem (Finds largest // power of P that divides x! function largestPower($x, $P) { // Calculate x/p + x/(p^2) + x/(p^3) + .... $res = 0; while ($x) { $x = floor($x/$P); $res += $x; } return $res; } // Returns sum of largest powers of P that divide // numbers in range from L to R. function sumOfDivisblePowers($L, $R, $P) { return largestPower($R, $P) - largestPower($L-1, $P); } // Driver code $L = 1; $R = 10; $P = 2; echo sumOfDivisblePowers($L, $R, $P); //This code is contributed by mits. ?>

**Output:**

8

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