using System;

namespace ProjectEuler_18
{
    internal class Program
    {
        private static int[][] GetGreatestPathSum(int[][] triangle, int index)
        {
            if (index < triangle.Length - 1)
            {
                triangle = GetGreatestPathSum(triangle, index + 1);
            }

            if (index - 1 < 0)
            {
                return triangle;
            }

            int[] top = triangle[index - 1];
            int[] bottom = triangle[index];
            int[] result = new int[top.Length];

            for (int i = 0; i < top.Length; i++)
            {
                result[i] = Math.Max(top[i] + bottom[i], top[i] + bottom[i + 1]);
            }

            triangle[index - 1] = result;

            return triangle;
        }

        private static void Main(string[] args)
        {
            string triangle =
@"75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23";

            string[] triangleLines = triangle.Split(new char[] { '\r', '\n' }, StringSplitOptions.RemoveEmptyEntries);
            int[][] intTriangle = new int[triangleLines.Length][];

            for (int i = 0; i < triangleLines.Length; i++)
            {
                string[] numberStrings = triangleLines[i].Split(' ');
                int[] numbers = new int[numberStrings.Length];

                for (int x = 0; x < numberStrings.Length; x++)
                {
                    numbers[x] = int.Parse(numberStrings[x]);
                }

                intTriangle[i] = numbers;
            }

            int sum = GetGreatestPathSum(intTriangle, 0)[0][0];

            Console.WriteLine("Greatest path sum: {0}", sum);
        }
    }
}

In this challenge, the goal is to calculate the number of lattice paths in a square grid. A lattice path is define as: A path composed of connected horizontal and vertical line segments, each passing between adjacent lattice points. They have a lovely image on the challenge site showing an example of lattice paths in a 2x2 grid.

With this problem, it helped that I have been working on some path finding algorithms for a game I'm developing. So I know that lattice paths can be easily calculated on a square grid by using Pascal's Triangles. The equation to find a node in Pascal's Triangle is: Where n is the row number in the triangle and k is the node number to select. In a Pascal Triangle, the lattice path number will be at 2 times the number of cells in the row and in the middle of the triangle. So for a 4x4 grid, the lattice paths count will be a n=8 and k=4.

using System;
using System.Numerics;

namespace ProjectEuler_15
{
    internal class Program
    {
        private static BigInteger Factorial(int n)
        {
            BigInteger b = 1;

            for (int i = n; i > 1; i--)
            {
                b *= i;
            }

            return b;
        }

        private static BigInteger GetLatticePathCount(int squareSize)
        {
            // lattice path counts of a square can be found using
            // pascal's triangle.  The number of lattice paths will
            // be the center value of pascal's triangle at the 2x row number
            // of the square size.
            // formula for pascal's triangle:
            //    n!
            // ________
            // k!(n-k)!

            int n = 2 * squareSize, k = squareSize;

            var result = Factorial(n) / (Factorial(k) * Factorial(n - k));

            return result;
        }

        private static void Main(string[] args)
        {
            var latticeCount = GetLatticePathCount(20);
            Console.WriteLine("Number of lattice paths in a 20x20 grid: {0}", latticeCount);
        }
    }
}

This challenge wants you to find the longest collatz series between 1 and 1000000. A collatz series is defined by:
n → n/2 (n is even)
n → 3n + 1 (n is odd)

using System;

namespace ProjectEuler_14
{
    internal class Program
    {
        private static void Main(string[] args)
        {
            const int MaxNum = 1000000;
            int highestCollatzSeriesCount = 0;
            int numWithHighestSeries = 0;
            // a cache can be used because all the series number eventually will get to
            // a series that's already been calculated; therefore, this will save a lot
            // of calculation time
            int[] cache = new int[MaxNum + 1];

            // 1 has a collatz series count of 1
            cache[1] = 1;

            // start the loop at 2 since, in theory, all numbers end with 1
            for (int i = 2; i < MaxNum; i++)
            {
                int count = 0;
                // use a long because int will likely overflow at higher numbers
                long collatz = i;
                do
                {
                    //n → n/2 (n is even)
                    //n → 3n + 1(n is odd)
                    if (collatz % 2 == 0)
                    {
                        collatz /= 2;
                    }
                    else
                    {
                        collatz = (collatz * 3) + 1;
                    }

                    count++;
                    // we can break the loop if collatz=1 or collatz > i
                    // because if it is less than i, it's been cached
                } while (collatz != 1 && collatz > i);

                // add the cached number to the total count
                count += cache[collatz];

                // cache the value
                cache[i] = count;

                if (count > highestCollatzSeriesCount)
                {
                    highestCollatzSeriesCount = count;
                    numWithHighestSeries = i;
                }
            }

            // print out the result
            Console.WriteLine("Number with highest collatz series below {0}: {1} with {2} in series.", MaxNum, numWithHighestSeries, cache[numWithHighestSeries]);
        }
    }
}

Since memory wasn't a huge concern, I decided to cache the collatz sequence counts for every calculation so that I am not wasting CPU cycles repeating calculations. This ended up making the total run time to be less than 1 second whereas without caching took roughly 30 seconds.

This challenge wants you to find the largest product in a 20x20 grid that is 4 numbers in all directions (up, down, left, right, and all diagonals). This was pretty straight forward, I went with a brute force method.

namespace ProjectEuler_11
{
    internal class Program
    {
        private static int[,] _data =
        {
            {08,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,08},
            {49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00},
            {81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65},
            {52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91},
            {22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},
            {24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50},
            {32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},
            {67,26,20,68,02,62,12,20,95,63,94,39,63,08,40,91,66,49,94,21},
            {24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
            {21,36,23,09,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95},
            {78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,09,53,56,92},
            {16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57},
            {86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58},
            {19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40},
            {04,52,08,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66},
            {88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69},
            {04,42,16,73,38,25,39,11,24,94,72,18,08,46,29,32,40,62,76,36},
            {20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16},
            {20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54},
            {01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48}
        };

        private static int GetLargerHorizontalProduct(int multipleCount, int[,] data, int x, int y, int rows, int cols, int largestProduct)
        {
            // make sure we have enough colums to get the number of values we need
            if (x + multipleCount >= cols)
            {
                return largestProduct;
            }

            int product = data[x, y];

            for (int i = x + 1; i < multipleCount + x; i++)
            {
                product *= data[i, y];
            }

            return product > largestProduct ? product : largestProduct;
        }

        private static int GetLargerLeftDiagnolProduct(int multipleCount, int[,] data, int x, int y, int rows, int cols, int largestProduct)
        {
            if (x + multipleCount >= cols || y - multipleCount < 0)
            {
                return largestProduct;
            }

            int product = data[x, y];

            for (int i = x + 1, j = y - 1; i < x + multipleCount; i++, j--)
            {
                product *= data[i, j];
            }

            return product > largestProduct ? product : largestProduct;
        }

        private static int GetLargerProduct(int multipleCount, int[,] data, int rows, int cols, int largestProduct, int x, int y)
        {
            largestProduct = GetLargerHorizontalProduct(multipleCount, data, x, y, rows, cols, largestProduct);
            largestProduct = GetLargerVerticalProduct(multipleCount, data, x, y, rows, cols, largestProduct);
            largestProduct = GetLargerRightDiagnolProduct(multipleCount, data, x, y, rows, cols, largestProduct);
            largestProduct = GetLargerLeftDiagnolProduct(multipleCount, data, x, y, rows, cols, largestProduct);
            return largestProduct;
        }

        private static int GetLargerRightDiagnolProduct(int multipleCount, int[,] data, int x, int y, int rows, int cols, int largestProduct)
        {
            // make sure we have enough room to get the number of values we need
            if (x + multipleCount >= cols || y + multipleCount >= rows)
            {
                return largestProduct;
            }

            int product = data[x, y];

            for (int i = x + 1, j = y + 1; i < x + multipleCount; i++, j++)
            {
                product *= data[i, j];
            }

            return product > largestProduct ? product : largestProduct;
        }

        private static int GetLargerVerticalProduct(int multipleCount, int[,] data, int x, int y, int rows, int cols, int largestProduct)
        {
            // make sure we have enough rows to get the number of values we need
            if (y + multipleCount >= rows)
            {
                return largestProduct;
            }

            int product = data[x, y];

            for (int i = y + 1; i < y + multipleCount; i++)
            {
                product *= data[x, i];
            }

            return product > largestProduct ? product : largestProduct;
        }

        private static void Main(string[] args)
        {
            const int multipleCount = 4;
            int rows = _data.GetLength(0);
            int cols = _data.GetLength(1);
            int largestProduct = 0;

            for (int x = 0; x < rows; x++)
            {
                for (int y = 0; y < cols; y++)
                {
                    largestProduct = GetLargerProduct(multipleCount, _data, rows, cols, largestProduct, x, y);
                }
            }

            System.Console.WriteLine("Largest multiple found: {0}", largestProduct);
        }
    }
}

This challenge wants you to return the first triangular number that has over 500 factors. This challenge was a 2 part, first you have to be able to calculate out a "triangular number" which they define as a natural number added upon itself (similar to Fibonacci numbers) so that the 7th number in the series would be 1+2+3+4+5+6+7 or 28. The second part is being able to quickly calculate all the factors of a number. This was relatively straight forward, so I'll just post the code, there are comments in certain areas as to why I chose to do it like I did.

using System;
using System.Collections.Generic;
using System.Diagnostics;

namespace ProjectEuler_12
{
    internal class Program
    {
        private static HashSet GetMultiples(long num)
        {
            // use a hashset to guarantee uniqueness
            HashSet multiples = new HashSet { 1L, num };

            long root = (long)Math.Sqrt(num);

            // we only need to loop up to the square root of the num
            // because we're getting both the low and high factor of the num at once
            // no number could be higher than root*root
            for (long i = 2; i <= root; i++)
            {
                if (num % i == 0)
                {
                    multiples.Add(i);
                    multiples.Add(num / i);
                }
            }

            return multiples;
        }

        private static void Main(string[] args)
        {
            IEnumerable triangularNumbers = TriangularNumberGenerator();

            int multipleCount = 0;
            using (var enumerator = triangularNumbers.GetEnumerator())
            {
                long count = 0;
                long highestMultipleCountAchieved = 0;

                while (multipleCount < 500)
                {
                    enumerator.MoveNext();
                    count++;

                    multipleCount = GetMultiples(enumerator.Current).Count;

                    if (multipleCount > highestMultipleCountAchieved)
                    {
                        highestMultipleCountAchieved = multipleCount;
                        Debug.WriteLine("Number: {2}, Index: {0}, Multiple Count: {1}", count, highestMultipleCountAchieved, enumerator.Current);
                    }
                }

                Console.WriteLine("First triangular number index to have 500 multiples: {0}\r\nTriangular Number: {1}", count, enumerator.Current);
            }
        }

        /// <summary>
        /// The sequence of triangle numbers is generated by adding the natural numbers.
        /// </summary>
        /// <returns></returns>
        private static IEnumerable TriangularNumberGenerator()
        {
            long prevNum = 0;
            long sum = 0;

            while (true)
            {
                sum += ++prevNum;

                // use of yield return lets us go up as high as we need and only
                // as high as we need in this infinite series
                yield return sum;
            }
        }
    }
}