public static long Solve()
{
long sum = 0;
long sumFirst = 0;
long sumSecond = 0;
for (int i = 1; i < 10000; i++)
{
sumFirst = 0;
sumSecond = 0;
// we get the divisors of i and add them together
long[] divisors = Toolbox.GetDivisors(i);
for (int j = 0; j < divisors.Length - 1; j++)
{
sumFirst += divisors[j];
}
// if the sum of divisors is not equal to the number itself we can continue
if (sumFirst != i)
{
// we get the divisors of the previous sum
divisors = Toolbox.GetDivisors(sumFirst);
for (int j = 0; j < divisors.Length - 1; j++)
{
sumSecond += divisors[j];
}
// if the second sum matches the original number, the number is considered to be amicable
if (sumSecond == i)
{
sum += sumFirst + i;
}
}
}
// we divide the result by 2, because we added each number of the pair twice
return(sum / 2);
}
public static long Solve()
{
// first we need to find all the abundant numbers up to cca. 28123 - just to be safe
long[] divisors;
long sum = 0;
List <int> abundant = new List <int>();
for (int i = 1; i < 28123; i++)
{
sum = 0;
divisors = Toolbox.GetDivisors(i);
for (int j = 0; j < divisors.Length - 1; j++)
{
sum += divisors[j];
}
if (sum > i)
{
abundant.Add(i);
}
}
// then we need to add each number in the set to each other to get all the possible numbers
int first, second, third;
List <int> validNumbers = new List <int>();
for (int i = 0; i < abundant.Count; i++)
{
first = abundant.ElementAt(i);
for (int j = 0; j < abundant.Count; j++)
{
second = abundant.ElementAt(j);
third = first + second;
if (third < 28123)
{
validNumbers.Add(third);
}
}
}
validNumbers = validNumbers.Distinct().ToList();
sum = 0;
// finally we need to check which numbers are not contained in the set and add them together
for (int i = 1; i < 28123; i++)
{
if (!validNumbers.Contains(i))
{
sum += i;
}
}
return(sum);
}
public static long Solve()
{
long[] divisors = new long[] { };
int num = 1;
int counter = 2;
while (divisors.Length <= 500)
{
num += counter;
counter++;
divisors = Toolbox.GetDivisors(num);
}
return(num);
}
