JavaScript Program to Print All Numbers Divisible by 3 and 5 of a Given Number

In this article, we are going to discuss how to print all the numbers that are divisible by 3 and 5. We will write a program for finding out those numbers. For this, we can use loop or conditional statements.

Example:

Input: 75
Output: 0, 15, 30, 45, 60, 75
Input: 150
Output: 0, 15, 30, 45, 60, 75, 90, 105, 120, 135, 150

We can solve this by using the following methods:

Approach 1: Using for Loop

Using n = 50 as an example, the program should output all numbers below 50 that are divisible by both 3 and 5. To do this, divide each number between 0 and n by 3 and 5 in a for loop. Print that value if the remainder in both scenarios is 0.

Example: This example describes how can we print numbers that are divisible by 3 and 5 using a for loop.

// Function for finding out 
// the numbers which are 
// divisible by 3 and 5
function divisibleBy3and5(n) {
    for (let i = 0; i <= n; i++) {
        if (i % 3 == 0 && i % 5 == 0) {
            console.log(i)
        }
    }
}

let N = 50;

// Calling function for printing 
// the numbers
divisibleBy3and5(N);

0
15
30
45

Time Complexity: It is taking O(N) time as for loop will iterate n (given number) times.

Auxiliary Space: The space complexity of the above code is O(1), as no extra space is required in order to complete the operation.

Approach 2: Using LCM Method

Here we are taking LCM of 3 anad 5 which is 15. If any number which is divisible by 15 it means it can also divisible by 3 and 5 so we are only searching for that number which is divisible by 15 and printing that number only.

Example: This example is showing how can you print number which are divisible by 3 and 5 using LCM

// Creating variable having 
// value 50
let n = 50;

// For loop for printing
// numbers which are divisible
// by both 3 and 5
for (let i = 0; i <= 50; i++) {

    // Lcm of 3 and 5 is 15
    if (i % 15 == 0) {
        console.log(i);
    }
}

0
15
30
45

Time Complexity: The time complexity will be O(n) as loop will iterate n times.

Auxiliary Space: The space complexity of the above code is O(1), as no extra space is required in order to complete the operation.

Approach 3: Using the Arithmetic Series Formula

In this approach, we are going to iterate everytime with the increment of 15, as 15 is the LCM of 3 and 5, And if number is divisible by 15 then it will also be divisible by 3 and 5. And it will only work if we are starting our iteration from 0.

Example: This example describes how you can find out the numbers which are divisible by 3 and 5 using arithmetic series formula

// JavaScript program to print 
// numbers that are divisible
// by 3 and 5

// Lcm of 3 and 5 is 15
// LCM(3, 5) = 15
let n = 150;

// Function for finding 
// out divisible number

function divisibility(n) {

    // Start loop from 0 to n
    // by increment of +15 every time
    for (let i = 0; i <= n; i += 15) {
        console.log(i);
    }
}

// Calling function divisibility
divisibility(n);

0
15
30
45
60
75
90
105
120
135
150

Time Complexity: The time complexity will be O(n) as loop will iterate n times.

Auxiliary Space: The space complexity of the above code is O(1), as no extra space is required in order to complete the operation.

Approach 4: Using a while loop

We can iterate through the numbers starting from 0 and increment by 1 until the given number (inclusive). For each number, we check if it is divisible by both 3 and 5. If it is, we print it.

Example:

function printDivisibleBy3And5(n) {
    let num = 0;
    while (num <= n) {
        if (num % 3 === 0 && num % 5 === 0) {
            console.log(num);
        }
        num++;
    }
}

// Test cases
printDivisibleBy3And5(75);
printDivisibleBy3And5(150);

0
15
30
45
60
75
0
15
30
45
60
75
90
105
120
135
150