Question

How to check if a number is abundant number in JavaScript

Answer 1

function SumNumberProperDivisors(num) {
    let sum = 0;
    
    for (let i = 1; i <= Math.sqrt(num); i++) {
        if (num % i == 0) {
            if (i == (parseInt(num / i))) {
                sum += i;
                console.log(i + ", ");
            }
            else {
                sum += i + (parseInt(num / i));
                console.log(i + ", " + parseInt(num / i) + ", ");
            }
        }
    }
    return sum - num;
}
        
let num = 24;
let sum = SumNumberProperDivisors(num);
console.log("num = " + num + " sum = " + sum + " abundant = " + (sum > num ? "true" : "false"));
        
num = 21;
sum = SumNumberProperDivisors(num);
console.log("num = " + num + " sum = " + sum + " abundant = " + (sum > num ? "true" : "false"));





/*
run:

"1, 24, "
"2, 12, "
"3, 8, "
"4, 6, "
"num = 24 sum = 36 abundant = true"
"1, 21, "
"3, 7, "
"num = 21 sum = 11 abundant = false"

*/

 

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