Suppose you're talking about N factorial. For an example, we'll let N be 20.
For each Prime number p, repeatedly divide N by p, discarding the remainder. Add up the answers.
20/2 = 10, 10/2 = 5, 5/2 = 2, 2/2 = 1. 10+5+2+1 = 18.
20/3 = 6, 6/3 = 2. 6+2 = 8.
20/5 = 4
20/7 = 2
20/11 = 1
20/13 = 1
20/17 = 1
20/19 = 1
Now add 1 to each of these results: 19,9,5,3,2,2,2,2.
Now multiply these numbers. 19x9x5x3x2x2x2x2 = 41040.
This is how many factors the factorial has. 20! has 41040 factors. (This includes both 1 and the factorial itself. Subtract 1 if you're only counting the proper factors.)
The proof is left as an exercise for the reader.
To calculate the number of zeros in a factorial number, we need to determine the number of factors of 5 in the factorial. In this case, we are looking at 10 to the power of 10 factorial. The number of factors of 5 in 10! is 2 (from 5 and 10). Therefore, the number of zeros in 10 to the power of 10 factorial would be 2.
In Prolog, a simple factorial program can be defined using recursion. Here's a basic implementation: factorial(0, 1). % Base case: factorial of 0 is 1 factorial(N, Result) :- N > 0, N1 is N - 1, factorial(N1, Result1), Result is N * Result1. % Recursive case You can query the factorial of a number by calling factorial(N, Result). where N is the number you want to compute the factorial for.
It is impossible to determine what number has the most factors because there are an infinite number of numbers.
The factorial of a number is the product of all the whole numbers, except zero, that are less than or equal to that number.
a factorial number is a number multiplied by all the positive integers i.e. 4!=1x2x3x4=24 pi!=0.14x1.14x2.14x3.14 0!=1
To calculate the number of zeros in a factorial number, we need to determine the number of factors of 5 in the factorial. In this case, we are looking at 10 to the power of 10 factorial. The number of factors of 5 in 10! is 2 (from 5 and 10). Therefore, the number of zeros in 10 to the power of 10 factorial would be 2.
1,2,3,4,5
270
#include <iostream> using namespace std; int main() { int i, number=0, factorial=1; // User input must be an integer number between 1 and 10 while(number<1 number>10) { cout << "Enter integer number (1-10) = "; cin >> number; } // Calculate the factorial with a FOR loop for(i=1; i<=number; i++) { factorial = factorial*i; } // Output result cout << "Factorial = " << factorial << endl;
The factorial of 569 is 569 * 568 * 567 * ... * 3 * 2 * 1. (That is a very, very large number.) If you mean, however, "what are the factors of 569?", the answer is 1 and 569, because 569 is prime.
The time complexity for calculating the factorial of a number is O(n), where n is the number for which the factorial is being calculated.
double factorial(double N){double total = 1;while (N > 1){total *= N;N--;}return total; // We are returning the value in variable title total//return factorial;}int main(){double myNumber = 0;cout > myNumber;cout
Pseudo code+factorial
In Prolog, a simple factorial program can be defined using recursion. Here's a basic implementation: factorial(0, 1). % Base case: factorial of 0 is 1 factorial(N, Result) :- N > 0, N1 is N - 1, factorial(N1, Result1), Result is N * Result1. % Recursive case You can query the factorial of a number by calling factorial(N, Result). where N is the number you want to compute the factorial for.
A big number.
It is impossible to determine what number has the most factors because there are an infinite number of numbers.
The factorial of a number is the product of all the whole numbers, except zero, that are less than or equal to that number.