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Editorial Team
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Asked: 2026-05-27T01:33:36+00:00

I’m stuck with the JavaScript code I am using for solving a problem which

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I’m stuck with the JavaScript code I am using for solving a problem which states:

The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?

(this is not homework, is an online coding/mathematical challenge)

So I came up with this solution:

<html>
    <head>
    <script type="text/javascript">

        // This function checks whether it's possible to divide the prime number
        function al(n){
            // k = 13195 is the number that I have to find the prime factor for
            var k = 13195;
            if (k%n) {
                return;
            }
            else {
                document.write(n + '   ');
            }
        }
    </script>
    </head>
    <body>
    <script type="text/javascript">

        //a and b are just for counting from the number n to 2 to find the prime numbers
        var a = 2;
        for (var n = 13194 ; n>a ; n--) {
            var b = 2;
            //if the found number is divisible, we skip it
            while(b<n) {
                if (n % b == 0) {
                    break;
                }
                else if (b = n - 1){
                    al(n);
                }
                b++;
            }
        }
    </script>
    </body>
</html>

Now to the problem: This outputs these numbers: “2639 1885 1015 455 377 203 145 91 65 35 29 13 7 5”. These aren’t prime numbers. I’ve looked at the numbers and I found that the number 13195 divided by 5 (the last number) gives the first number; 2639; 13195 divided by 7 gives 1885; etc.

What exactly am I doing wrong?

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1 Answer

  1. Editorial Team

    Your issue isn’t a mathematical one — you just have a bug in one of your conditional checks.

    Change if(b = n-1) to if(b == n-1). Right now, your code isn’t actually checking to ensure that a factor is prime; instead, it’s assigning the value of n-1 to b (for odd values of n) and then automatically calling a1(b), so your result is all possible odd factors of k.

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