Problem #23 says:

Let p(n) be the nth prime: 2, 3, 5, 7, 11, …, and let r be the remainder when \((p(n)-1)^{n}+(p(n)+1)^{n}\) is divided by p(n)2.

For example, when n = 3, p(3) = 5, and 43 + 63 = 280 ≡ 5 mod 25.

The least value of n for which the remainder first exceeds \(10^{9}\) is 7037.

Find the least value of n for which the remainder first exceeds \(10^{10}\).

This can be solved using a modified version of the solution for Problem #120. We don’t even need to keep track of all of the remainders because the problem only concerns itself with the first remainder to exceed \(10^{10}\). So using our maximum remainder from the previous Problem #120 which was \(2*n*a\) we can modify it to be \(2*p_{n}*n\). From this point we just need a collection of prime numbers large enough to get the Job done; I chose to use a list of primes going up to 300000 because this creates a list of 25998 prime numbers and \(25998 * 2 * p[25998] > 1.5 x 10^{10}\) which gives me all of the data I needed to find a solution exceeding \(10^{10}\).

Solution below in c#/mono. runs in about .113 seconds.

using System;
using System.Collections.Generic;

namespace Problem123
{
	class MainClass
	{
		public static void Main (string[] args)
		{
			ulong upperbound = 10000000000ul;
			List<ulong> p = GeneratePrimes(300000);
			p.Insert(0,0UL);
			Console.WriteLine("primes done");
			for (uint n = 1; n < 300000ul; n++)
			{
				if (n%2 ==0ul)
					continue;
				ulong r = 2 * p[(int)n] * n;
				if (r > upperbound)
				{
					Console.WriteLine(n);
					Console.WriteLine(p[(int)n]);
					Console.WriteLine(r);
					break;
				}
			}

		}
		static List<ulong> GeneratePrimes(ulong num)
		{
			List<ulong> RetValue = new List<ulong>();
			RetValue.Add((ulong)2);
			RetValue.Add((ulong)3);
			ulong Stepper = 5;
			ulong Check = 1;
			while (Stepper <= num)
			{
				foreach (ulong i in RetValue)
				{
					if (Stepper % i == 0)
					{
						Check = 0;
						break;
					}
					if (Math.Sqrt(Stepper) <= i)
					{
						break;
					}
				}
				if (Check == 1)
				{
					//Console.WriteLine(((float)Stepper / (float)num) * 100);
					RetValue.Add(Stepper);
				}
				Check = 1;
				Stepper++;
				Stepper++;
			}
			return RetValue;
		}
	}
}

Also just as a note I’ve moved from jsMath to MathJax to see if it solves any of the issues I’ve been having. If you see something that might be an equation but it’s all screwy please point it out in the comments.

Problem #86 says:

A spider, S, sits in one corner of a cuboid room, measuring 6 by 5 by 3, and a fly, F, sits in the opposite corner. By travelling on the surfaces of the room the shortest “straight line” distance from S to F is 10 and the path is shown on the diagram.

However, there are up to three “shortest” path candidates for any given cuboid and the shortest route is not always integer.

By considering all cuboid rooms with integer dimensions, up to a maximum size of M by M by M, there are exactly 2060 cuboids for which the shortest distance is integer when M=100, and this is the least value of M for which the number of solutions first exceeds two thousand; the number of solutions is 1975 when M=99.

Find the least value of M such that the number of solutions first exceeds one million.

I thought this problem was going to be more interesting than it turned out being. Basically we’re looking for triplets (a,b,M) where 1<=a<=b<=M that make the equation \((a+b)^{2}+n^{2}\) a perfect square. We just do this for every value M=1-?? and count the number of solutions for that M value until the sum of all previous solution counts equals one million. Basically we take M = 3 which has a solution count of 3 and is the first none zero solution count and M = 4 with a solution count of 1; We add the solutions counts for M = 1-4 which are {0,0,2,1} and check to see if the sum of that series is greater than one million. If that sum isn’t one million we calculate the next value of M and add that to the series and check the series sum again; wash, rinse, repeat.

This problem became a lot less interesting when I found the formula for the shortest point and googled it and came back with two OEIS sequences (A143714,A143715) which made the solution trivial to write because it showed the method of building on the previous M values. On the other hand trivial to program solutions mean a lot less time spent frustrated trying to get the program to work.

Solution provided in c#/mono and runs in ~30ish seconds. Not the fastest but it works.

using System;
using System.Linq;
using System.Collections.Generic;

namespace Problem86
{
	class MainClass
	{
		public static List<int> A143714 = new List<int>();

		public static void Main (string[] args)
		{
			A143714.Add(0);
			A143714.Add(0);

			while(A143714.Sum() < 1000000)
			{
				int n = A143714.Count+1;

				int count = 0;
				for (int b = 1;b<=n;b++)
				{
					for (int a = 1;a<=b;a++)
					{
						if (IsPerfectSquare(((a+b)*(a+b))+(n*n)))
							count++;
					}
				}

				A143714.Add(count);
			}
			Console.WriteLine(A143714.Count);
		}

		public static bool IsPerfectSquare(int input)
		{
		    var sqrt = Math.Sqrt((double)input);
		    return Math.Ceiling(sqrt) == Math.Floor(sqrt);
		}

	}
}

With this solution I have only 11 problems left between 1-100 and only 39 left until I hit level 4!

Problem #80 is stated as:

It is well known that if the square root of a natural number is not an integer, then it is irrational. The decimal expansion of such square roots is infinite without any repeating pattern at all.

The square root of two is 1.41421356237309504880…, and the digital sum of the first one hundred decimal digits is 475.

For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits for all the irrational square roots.

This problem actually lead me to a new and interesting way of finding square roots described in a paper titled “Square roots by subtraction” by Frazer Jarvis. Frazer claims that it is an old math algorithm from Japan but it’s not the algorithms history that makes it useful. Instead it’s the fact that the algorithm allows us to get arbitrarily precise square roots using basic subtraction and multiplication (I felt the title was a little odd but oh well).

The algorithm works like this take your number \(n\) and make a new number \(a=5n\) and let \(b=5\). The you apply one of two rules depending on whether \(a\) is larger than \(b\). If \(a\) is larger then you apply Rule 1: which is to set \(a=a-b\) and set \(b=b+10\). If \(b\) is larger you apply Rule 2: set \(a= a*100\) and set \(b=((b-5)*10)+5)\). Just repeat those rules using the \((a,b)\) from the previous round until your answer (which is \(b\)) has the desired precision.

So for the square root of 2 we get (10,5) \(Rule 1\atop \longrightarrow\) (5,15) \(Rule 2\atop \longrightarrow\) (500,105)\(Rule 1\atop \longrightarrow\) (395,115) \(Rule 1\atop \longrightarrow\) (280,125) \(Rule 1\atop \longrightarrow\) (155,135) \(Rule 1\atop \longrightarrow\) (20,145) \(Rule 2\atop \longrightarrow\) (2000,1405) \(Rule 1\atop \longrightarrow\) (595,1415) and as we continue b gets closer and closer to the square root of 2. With this algorithm we are only limited by the size of the integer class we are using. As an added bonus this algorithm uses an integer class so we don’t have to worry about rounding errors. This means that with c# and .Net 4 we can using the BigInteger class from System.Numerics and generate the digits of the square root of 2 forever*.

So once we have this method all figured out what’s left to do? Well as it turns out not a whole lot; we just need a digital sum and a list of perfect squares to exclude and we’re golden. Solution provided in c#/mono and runs in 1 second or so. Could easily be made faster (at the very least one could add some parallelism).

using System;
using System.Numerics;
using System.Collections.Generic;

namespace Problem80
{
	class MainClass
	{
		public static void Main (string[] args)
		{
			List<int> squares = new List<int>();
			for (int i = 1; i <= 10; i++)
			{
				squares.Add(i*i);
			}
			uint sum = 0;
			for (int i = 1; i <=100; i++)
			{
				if (!squares.Contains(i))
					sum += DigitalSum(i);
			}
			Console.WriteLine(sum);

		}
		public static uint DigitalSum (int i)
		{
			BigInteger digits = BigInteger.Parse(SquareRoot(i).ToString().Substring(0,100));
			uint digitsum = 0;
			while (digits != BigInteger.Zero)
			{
				uint temp = (uint)(digits%10);
				digitsum += temp;
				digits = (digits -temp) / 10;
			}

			return digitsum;
		}
		public static BigInteger SquareRoot (int i)
		{
			BigInteger a = 5 * i;
			BigInteger b = 5;
			while (b.ToString().Length < 102)
			{
				if (a >= b)
				{
					a = a-b;
					b += 10;
				}
				else
				{
					a *= 100;
					b = ((b-5)*10)+5;
				}
			}
			return b;
		}
	}
}

* Memory limits and the physics of our universe prevent a person or computer from actually doing this.

Problem #95 says:

The proper divisors of a number are all the divisors excluding the number itself. For example, the proper divisors of 28 are 1, 2, 4, 7, and 14. As the sum of these divisors is equal to 28, we call it a perfect number.

Interestingly the sum of the proper divisors of 220 is 284 and the sum of the proper divisors of 284 is 220, forming a chain of two numbers. For this reason, 220 and 284 are called an amicable pair.

Perhaps less well known are longer chains. For example, starting with 12496, we form a chain of five numbers:

12496 → 14288 → 15472 → 14536 → 14264 (→ 12496 → …)

Since this chain returns to its starting point, it is called an amicable chain.

Find the smallest member of the longest amicable chain with no element exceeding one million.

An extremely easy problem to solve despite the fact that it’s quite far down the list when sorted by difficulty. We can instantly stop generating the chain if it exceeds one million at any point in the chain. Also we need to check to make sure the chain ends up where we started or it isn’t valid for the problem; this means that we need to make sure we don’t get stuck in an infinite loop where the chain starts over and we miss it. Solution is written using a simple sum of divisors function that simply brute forces the sum instead of a more elaborate prime factor function; I don’t think the prime factor version would be too much faster because the magnitude of the numbers (under one million) just isn’t enough to warrant such an effort.

Solution written in mono/c# with a helping hand from the linq name-space (mainly for list.min and list.max functions without having to write my own) runs in about 6 seconds on a modern machine. Interestingly enough this could be made parallel if one so desired with a few simple modifications but it runs fast enough already.

using System;
using System.Collections.Generic;
using System.Linq;

namespace Problem95
{
	class MainClass
	{
		public static void Main (string[] args)
		{

			List<long> chain = new List<long>();
			List<long> tmp = new List<long>();

			for (long i = 1; i <= 100000; i++)
			{
				tmp = GenerateChain(i);
				if (tmp.Count() > chain.Count())
				{
					List<long> ch = tmp.Where(x=> x > 1000000).ToList();
					if (ch.Count() == 0)
						chain = GenerateChain(i);
				}
			}
			Console.WriteLine(chain[0].ToString() + " :: " + chain.Min().ToString());
		}
		public static List<long> GenerateChain (long num)
		{
			List<long> chain = new List<long>();
			long previous = num;
			while (!chain.Contains(previous) && previous < 1000000L)
			{
				chain.Add(previous);

				previous = NextInChain(previous);
			}

			if (previous != chain[0])
				chain.Clear();

			if (previous > 1000000L)
				chain.Add(9999999L);

			return chain;
		}
		public static long NextInChain (long num)
		{
			return ProperDivisors(num).Sum();
		}
		public static List<long> ProperDivisors (long num)
		{
			long max = (long)Math.Ceiling(Math.Sqrt((double)num));

			List<long> RetValue = new List<long>();
			RetValue.Add(1L);
			if (num % max == 0d)
				RetValue.Add(max);

			for(long i = 2; i < max; i++)
			{
				if (num % i == 0L)
				{
					RetValue.Add(i);
					RetValue.Add(num/i);
				}
			}

			return RetValue.Distinct().ToList();
		}
	}
}

Problem #61 asks

Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.

There are several things to look out for when writing a solution to this problem. 1) If the last two digits of a number are less than ten it cannot be part of a chain (because the question doesn’t allow numbers of the form 0XXX). 2) it doesn’t matter what type of number you start out with: because it’s a cyclical chain if you find one number you’ll find the others. 4) Don’t throw out duplicate numbers if they only differ by type: just because a triangle number and a heptagonal number are the same doesn’t mean that you can throw one away they can be valid parts of different chains. 4) Recursion.

Because we are told that the chain of numbers we’re looking for is cyclical a recursive approach is perhaps the easiest to write. We use a tiny class to keep track of the first two digits, the last two digits, the type of number (tri,hep,oct etc) and lastly the value of the number itself (so we can calculate the sum of the chain later). We generate all numbers of all types between 1,000 and 10,000 using this class and put them into a list. We then strip invalid numbers from the list; those with ending digits less than 10 so if a number ends with 09 then it cannot be part of a valid chain. Then we just loop through this giant list. To get the next number in the chain we make sure we don’t already have a number of that type in the chain and we don’t have a number with that value in the chain. If the number is a valid new addition to the solution chain we attempt to find the next number to the chain using the same rules. If a chain cannot be found with 6 members or the required cyclical property then we go up a level and remove the last number added and continue to check for a valid chain. Eventually using this solution we’ll find the chain and if we don’t stop with the first chain we’ll get all permutations of the chain; still we only need the first one because the sum of the numbers in the chain are all going to be the same regardless of the order they appear in.

Solution written in mono/c# and runs in less than 1/10th of a second.

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

namespace Problem61
{
	public class ChainNum
	{
		public int starts = 0;
		public int ends = 0;
		public int num = 0;
		public int type = 0;

		public ChainNum (int i,int t)
		{
			num = i;
			starts = i/100;
			ends = i % 100;
			type = t;
		}

	}
	class MainClass
	{
		public static List<ChainNum> Nums = new List<ChainNum>();

		public static Stopwatch sw = new Stopwatch();
		public static void Main (string[] args)
		{
			sw.Start();
			FillLists();
			List<ChainNum> inv = new List<ChainNum>();
			foreach (ChainNum c in Nums)
			{
				if (c.ends < 10)
					inv.Add(c);
			}
			foreach (ChainNum c in inv)
			{
				Nums.Remove(c);
			}
			foreach(ChainNum n in Nums)
			{
				var c = new List<ChainNum>();
				c.Add(n);
				var sol = BuildSolution(1,c);
				int count = 0;
				if (sol != null)
				{
					foreach (ChainNum s in sol)
						count += s.num;
				}
				if (count != 0)
				{
					Console.WriteLine(count);
					break;
				}
			}
			sw.Stop();
			Console.WriteLine(sw.Elapsed);
		}
		public static List<ChainNum> BuildSolution (int level, List<ChainNum> solution)
		{
			if (level > 6)
				return null;
			if (solution.Count == 6 && solution[0].starts == solution[5].ends)
				return solution;
			if (solution.Count == 6)
				return null;

			foreach (ChainNum c in Nums)
			{
				bool alreadyhastypeornumber = false;
				foreach(ChainNum temp in solution)
				{
					if (temp.type == c.type || temp.num == c.num)
						alreadyhastypeornumber = true;
				}
				if (alreadyhastypeornumber)
					continue;
				else
				{
					if (solution[solution.Count-1].ends == c.starts)
					{
						solution.Add(c);
						var sol = BuildSolution(level++,solution);
						if (sol != null)
							return sol;
						else
							solution.Remove(c);

					}
				}

			}
			return null;
		}
		public static void FillLists ()
		{
			int i = 1;
			while(TriangleNum(i) < 10000)
			{
				if (TriangleNum(i) > 1000)
					Nums.Add(new ChainNum(TriangleNum(i),3));
				i++;
			}
			i = 0;

			while(SquareNum(i) < 10000)
			{
				if (SquareNum(i) > 1000)
					Nums.Add(new ChainNum(SquareNum(i),4));
				i++;
			}
			i = 0;

			while(PentagonalNum(i) < 10000)
			{
				if (PentagonalNum(i) > 1000)
					Nums.Add(new ChainNum(PentagonalNum(i),5));
				i++;
			}
			i = 0;

			while(HexagonalNum(i) < 10000)
			{
				if (HexagonalNum(i) > 1000)
					Nums.Add(new ChainNum(HexagonalNum(i),6));
				i++;
			}
			i = 0;

			while(HeptagonalNum(i) < 10000)
			{
				if (HeptagonalNum(i) > 1000)
					Nums.Add(new ChainNum(HeptagonalNum(i),7));
				i++;
			}
			i = 0;

			while(OctagonalNum(i) < 10000)
			{
				if (OctagonalNum(i) > 1000)
					Nums.Add(new ChainNum(OctagonalNum(i),8));
				i++;
			}

		}

		public static int TriangleNum (int n)
		{
			return (n*(n+1))/2;
		}
		public static int SquareNum (int n)
		{
			return n*n;
		}
		public static int PentagonalNum (int n)
		{
			return (n*(3*n-1))/2;
		}
		public static int HexagonalNum (int n)
		{
			return n*(2*n-1);
		}
		public static int HeptagonalNum (int n)
		{
			return (n*(5*n -3))/2;
		}
		public static int OctagonalNum (int n)
		{
			return n*(3*n -2);
		}

	}
}