Sum of all multiples in JavaScript

We need to write a JavaScript function that takes a number n as the first argument, followed by any number of divisor arguments. The function should sum all numbers up to n that are divisible by any of the specified divisors.

Problem Statement

Given a number n and multiple divisors, find the sum of all numbers from 1 to n that are multiples of any of the divisors.

For example, if we call:

sumMultiples(15, 2, 3)

We need to find numbers up to 15 that are divisible by either 2 or 3:

2, 3, 4, 6, 8, 9, 10, 12, 14, 15

The sum of these numbers is 83.

Solution

Here's the implementation using a while loop and rest parameters:

const num = 15;

const sumMultiple = (num, ...arr) => {
    const dividesAny = num => arr.some(el => num % el === 0);
    let sum = 0;
    while (num) {
        if (dividesAny(num)) {
            sum += num;
        }
        num--;
    }
    return sum;
};

console.log(sumMultiple(num, 2, 3));

83

How It Works

The function uses several key concepts:

• Rest Parameters: ...arr collects all divisor arguments into an array
• Helper Function: dividesAny() checks if a number is divisible by any divisor using some()
• While Loop: Iterates from n down to 1, checking each number
• Modulo Operator: num % el === 0 tests for exact divisibility

Alternative Approach with For Loop

const sumMultiplesFor = (num, ...divisors) => {
    let sum = 0;
    for (let i = 1; i <= num; i++) {
        if (divisors.some(divisor => i % divisor === 0)) {
            sum += i;
        }
    }
    return sum;
};

console.log(sumMultiplesFor(15, 2, 3));
console.log(sumMultiplesFor(10, 3, 5));

83
18

Testing with Multiple Examples

// Test various combinations
console.log("Sum of multiples of 2,3 up to 15:", sumMultiple(15, 2, 3));
console.log("Sum of multiples of 3,5 up to 10:", sumMultiple(10, 3, 5));
console.log("Sum of multiples of 7 up to 20:", sumMultiple(20, 7));
console.log("Sum of multiples of 2,5 up to 12:", sumMultiple(12, 2, 5));

Sum of multiples of 2,3 up to 15: 83
Sum of multiples of 3,5 up to 10: 18
Sum of multiples of 7 up to 20: 21
Sum of multiples of 2,5 up to 12: 42

Conclusion

This solution efficiently finds the sum of all multiples using rest parameters and the some() method. The approach works for any number of divisors and handles edge cases gracefully.