Re: 61. - THE SILVER CUBES

On Tue, 15 Jul 2025 8:19:38 +0000, David Entwistle wrote:

On Sat, 12 Jul 2025 23:14:47 -0000 (UTC), Richard Tobin wrote:
>

The first solution can be found in seconds with a fairly straighforward
program.  Obviously Dudeney didn't do that.

>
I tried the programmatic approach and didn't find a solution in seconds,
nor even minutes - I was probably doing a lot of unnecessary looping.
I've
looked at the solution and the common denominator has five digits.
>
The other path looks to be the way to go.

The C# program below finds the solution in under a minute:
using System;
namespace SumOfCubes
{
    class SumOfCubes
    {
        private const long MAX = 200000;
        static void Main(string[] args)
        {
            if (args.Length == 1 && long.TryParse(args[0], out long
number) && number > 0)
            {
                FindCubes(number);
            }
            else
            {
                Console.WriteLine("Usage: SumOfCubes <number>");
            }
        }
        private static void FindCubes(long number)
        {
            double third = 1.0 / 3.0;
            bool found = false;
            for (long a = 1; a < MAX && !found; a++)
            {
                long a3 = a * a * a;
                for (long b = a; b < MAX && !found; b++)
                {
                    long sum = a3 + b * b * b;
                    long c3 = sum / number;
                    if (c3 * number == sum)
                    {
                        long c = (long)Math.Round(Math.Pow(c3, third));
                        if (c * c * c == c3 && GCD(a, c) == 1 && GCD(b,
c) == 1)
                        {
                            Console.WriteLine($"({a} / {c})^3 + ({b} /
{c})^3 = {number}");
                            found = true;
                        }
                    }
                }
            }
        }
        private static long GCD(long number1, long number2)
        {
            long remainder;
            while (number2 != 0)
            {
                remainder = number1 % number2;
                number1 = number2;
                number2 = remainder;
            }
            return number1;
        }
    }
}
--