JavaScript: Number of trailing zeroes in a factorial

JavaScript Math: Exercise-96 with Solution

Factorial Trailing Zeroes

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:

JavaScript Code:

/**
 * Function to count the number of trailing zeroes in the factorial of a given number.
 * @param {number} n - The input number.
 * @returns {number} - The count of trailing zeroes.
 */
function test(n) {
    ctr = 0; // Initialize the counter for trailing zeroes
    i = 5; // Initialize the divisor
    while (n / i >= 1) {
        ctr += Math.floor(n / i); // Increment the counter based on the quotient
        i *= 5; // Update the divisor for the next iteration
    }
    return ctr; // Return the count of trailing zeroes
}
// Test cases
let n = 4;
console.log("n = " + n);
console.log("Number of trailing zeroes in the factorial: " + test(n));

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

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

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

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

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:

Live Demo:

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

Sample Solution-2:

JavaScript Code:

/**
 * Function to count the number of trailing zeroes in the factorial of a given number.
 * @param {number} n - The input number.
 * @returns {number} - The count of trailing zeroes.
 */
function test(n) {
    pow = Math.log(n) / Math.log(5); // Calculate the maximum power of 5 that divides n
    result = 0; // Initialize the result variable
    for (i = 1; i <= pow; i++) { // Iterate from 1 to the calculated power
        result += Math.floor(n / Math.pow(5, i)); // Update the result by summing up the quotients of n divided by powers of 5
    };
    return result; // Return the count of trailing zeroes
}

// Test cases
let 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));

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:

Live Demo:

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

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