How many zeros are there at the end of 50! (50! Is read as the factorial of 50, which means 50! = 50 * 49 *... * 3 * 2 * 1)

How many zeros are there at the end of 50! (50! Is read as the factorial of 50, which means 50! = 50 * 49 *... * 3 * 2 * 1)

2*5=10
The number of zeros is determined by the number of two and five
In continuous natural numbers, the number of 2 is much more than that of 5
So 50! There are as many 5's as there are 0's
Every five numbers has a factor of five. Every 25 numbers has a factor of 25
Where 25 = 5 * 5, 5 has been calculated for every 5 numbers, so the factor of 1 25 is only 1 5 more
50! Together, there are 50 / 5 + 50 / 25 = 10 + 2 = 12 5
There are 12 consecutive zeros at the end of 50

How many zeros are there at the end of the factorial of N

The number of zeros at the end of the product depends on the number of two and the number of five in the factor. For factorial, every two numbers have at least one factor of two, so the factor of two is enough. The factor of five is relatively small, and at least five consecutive numbers can guarantee that there must be one

How to calculate factorial with calculator

If your calculator has factorial, it will have the sign x
If x! Is no longer on the key but on the empty line above the key, you need to press the number first, then shift, and then press the key under the empty item
Not every calculator can calculate factorial

0 is 24, factorial

(0!+0!+0!+0!)!=24
0!=1
4!=24

What are the factorials (!) of 10?

Factorial of 1: 1
Factorial of 2: 2
Factorial of 3: 6
Factorial of 4: 24
Factorial of 5: 120
Factorial of 6: 720
Factorial of 7: 5040
Factorial of 8: 40320
Factorial of 9: 362880
Factorial of 10: 3628800

Is the factorial of any number (except 0 and 1) even?

Yes, because at least one factor is 2. (n! = 2 * 3 *. * n)

If the defined function has a return value, the following statement about the function call is wrong: select an answer a. the function call can be used as an exclusive function
If a defined function has a return value, the following description of the function call is incorrect ()
Choose an answer
A. Function calls can exist as stand-alone statements
B. Function calls can be used as arguments to a function
C. Function calls can appear in expressions
D. Function calls can be assigned as lvalues

If it is C language, this question should choose D
If it is C + +, there is no answer to this question. Because there are reference types in C + +, if a function returns a reference type, it can be assigned as an lvalue

The number of pairs in which the values of any two different elements in array a [n] are equal to the values of MM is given, and then the statistical results are returned. If a [i] + [J] is equal to mm and I is not equal to j, it is a pair
According to the function prototype "void DD (int a [], int n, int mm), write the function definition, use double for loop to count the number of pairs whose values of any two different elements in array a [n] are equal to the value of mm, and then return the statistical results. If a [i] + [J] is equal to mm, and I is not equal to j, it is a pair

int DD(int a[],int n,int MM)
{
int count =0 ;
for(int i=0;i

The continuous function f (x) on [0,1] can be approximated by Bernstein polynomials. What about the continuous function g (x) on [a, b]

The continuous function g (x) on [a, b] can also be approximated by Bernstein polynomials
t=(x-a)/(b-a) x=(b-a)t+a
h(t)=g((b-a)t+a),g(x)=h((x-a)/(b-a))
H (T) is a continuous function on [0,1], which can be approximated by Bernstein polynomials. Then t = (x-a) / (B-A) is substituted into the Bernstein Polynomials of H (T), and the Bernstein polynomials approximation of G (x) is obtained

The recursive and recursive functions for calculating the value of Hermite polynomial HN (x) are given respectively
H0（x）=1
H1（x）=2x
HN (x) = 2xhn-1 (x) - 2 (n-1) HN-2 (x) (for x > 1)

Non recursive recursive functions: long fun1 (int n, int x) {long a, B, t; int i; if (n==0) return 1; if (n==1) return 2 * x; a = 1; b = 2 * x; for (I = 2; I1) return (2 * x * H (n-1, x) -2* (n-1) *h (n-2, x)); "this sentence will make the recurrence depth too large, when n is large