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JavaScript: Number of trailing zeroes in a factorial

JavaScript Math: Exercise-96 with Solution

Write a JavaScript program that calculates the factorial of a number and returns the number of trailing zeroes.

In mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. The factorial of n also equals the product of n with the next smaller factorial:
n! = n x (n-1) x (n-2) x (n-3)x.....x 3 x 2 x 1
= n x (n-1)!
For example 4! = 4 x 3! = 4 x 3 x 2 x 1 =24
The value of 0! is 1, according to the convention for an empty product.

Test Data:
(4) -> 0
(9) -> 1
(10) -> 2
(23) -> 4
(35) -> 8

Sample Solution-1:

HTML Code:

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <title>JavaScript program to Number of trailing zeroes in a factorial</title>
</head>
<body>

</body>
</html>

JavaScript Code:

function test(n) {
    ctr = 0
    i = 5
    while (n/i>=1){
        ctr += Math.floor(n/i)
        i *= 5
    }
    return ctr;
};
n = 4
console.log("n = " + n)
console.log("Number of trailing zeroes in the said factorial: "+test(n));
n = 9
console.log("n = " + n)
console.log("Number of trailing zeroes in the said factorial: "+test(n));
n = 10
console.log("n = " + n)
console.log("Number of trailing zeroes in the said factorial: "+test(n));
n = 23
console.log("n = " + n)
console.log("Number of trailing zeroes in the said factorial: "+test(n));
n = 35
console.log("n = " + n)
console.log("Number of trailing zeroes in the said factorial: "+test(n));

Sample Output:

n = 4
Number of trailing zeroes in the said factorial: 0
n = 9
Number of trailing zeroes in the said factorial: 1
n = 10
Number of trailing zeroes in the said factorial: 2
n = 23
Number of trailing zeroes in the said factorial: 4
n = 35
Number of trailing zeroes in the said factorial: 8

Flowchart:

JavaScript: Number of trailing zeroes in a factorial.

Live Demo:

See the Pen javascript-math-exercise-96 by w3resource (@w3resource) on CodePen.


Sample Solution-2:

JavaScript Code:

function test(n) 
   {
    pow = Math.log(n) / Math.log(5);
    result = 0;
    for (i = 1; i <= pow; i++) {
        result += Math.floor(n / Math.pow(5, i));
    };
    return result;
}

n = 4
console.log("n = " + n)
console.log("Number of trailing zeroes in the said factorial: "+test(n));
n = 9
console.log("n = " + n)
console.log("Number of trailing zeroes in the said factorial: "+test(n));
n = 10
console.log("n = " + n)
console.log("Number of trailing zeroes in the said factorial: "+test(n));
n = 23
console.log("n = " + n)
console.log("Number of trailing zeroes in the said factorial: "+test(n));
n = 35
console.log("n = " + n)
console.log("Number of trailing zeroes in the said factorial: "+test(n));

Sample Output:

n = 4
Number of trailing zeroes in the said factorial: 0
n = 9
Number of trailing zeroes in the said factorial: 1
n = 10
Number of trailing zeroes in the said factorial: 2
n = 23
Number of trailing zeroes in the said factorial: 4
n = 35
Number of trailing zeroes in the said factorial: 8

Flowchart:

JavaScript: Number of trailing zeroes in a factorial.

Live Demo:

See the Pen javascript-math-exercise-96-1 by w3resource (@w3resource) on CodePen.


Improve this sample solution and post your code through Disqus

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