Given an integer n, find the power of a given prime number(r) in n!
Input : n = 6 r = 3 Factorial of 6 is 720 = 2^4 * 3^2 *5 (prime factor 2,3,5) and power of 3 is 2 Output : 2 Input : n = 2000 r = 7 Output : 330
A simple method is to first calculate factorial of n, then factorize it to find the power of a prime number.
The above method can cause overflow for a slightly bigger numbers as factorial of a number is a big number. The idea is to consider prime factors of a factorial n.
Legendre Factorization of n!
For any prime number p and any integer n, let Vp(n) be the exponent of the largest power of p that divides n (that is, the p-adic valuation of n). Then
Vp(n) = summation of floor(n / p^i) and i goes from 1 to infinity
While the formula on the right side is an infinite sum, for any particular values of n and p it has only finitely many nonzero terms: for every i large enough that p^i > n, one has floor(n/p^i) = 0 . since, the sum is convergent.
Power of ‘r’ in n! = floor(n/r) + floor(n/r^2) + floor(n/r^3) + ....
Program to count power of a no. r in n! based on above formula.
- Finding power of prime number p in n!
- Find power of power under mod of a prime
- Largest power of k in n! (factorial) where k may not be prime
- Check if given number is a power of d where d is a power of 2
- Check if a prime number can be expressed as sum of two Prime Numbers
- Find coordinates of a prime number in a Prime Spiral
- Highest power of a number that divides other number
- Sum of digits of a given number to a given power
- Find whether a given number is a power of 4 or not
- Check if a number is a power of another number
- GCD of a number raised to some power and another number
- Find the super power of a given Number
- Number of digits in 2 raised to power n
- Given a HUGE number check if it's a power of two.
- Highest power of 2 less than or equal to given number
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Improved By : jit_t