Project Euler Solutions 1 - 25 (Java)

Project Euler Solutions 1 - 25 (Java)

Output

F:\Project Euler Solutions>javac ProjectEulerSolutions.java

F:\Project Euler Solutions>java ProjectEulerSolutions

Answer to problem 1:: 233168
in less than 1.136617 ms

Answer to problem 2:: 4613732
in less than 0.806384 ms

Answer to problem 3:: 6857
in less than 1.056405 ms

Answer to problem 4:: 906609
in less than 226.182888 ms

Answer to problem 5:: 232792560
in less than 0.165544 ms

Answer to problem 6:: 25164150
in less than 0.14421 ms

Answer to problem 7:: 104743
in less than 145.561179 ms

Answer to problem 8:: 23514624000
in less than 1.044885 ms

Answer to problem 9:: 31875000
in less than 1.922094 ms

Answer to problem 10:: 142913828922
in less than 2225.412344 ms

Answer to problem 11:: 70600674
in less than 0.330234 ms

Answer to problem 12:: 76576500
in less than 1008.353536 ms

Answer to problem 13:: 5537376230
in less than 83.219629 ms

Answer to problem 14:: 837799
in less than 974.053225 ms

Answer to problem 15:: 137846528820
in less than 0.353272 ms

Answer to problem 16:: 1366
in less than 1.48733 ms

Answer to problem 17:: 21124
in less than 0.756038 ms

Answer to problem 18:: 1074
in less than 0.414712 ms

Answer to problem 19:: 171
in less than 2.693065 ms

Answer to problem 20:: 648
in less than 2.233555 ms

Answer to problem 21:: 31626
in less than 382.265878 ms

Answer to problem 22:: 871198282
in less than 601.421283 ms

Answer to problem 23:: 4179871
in less than 1913.355161 ms

Answer to problem 24:: 2783915460
in less than 19.609625 ms

Answer to problem 25:: 4782
in less than 8.572838 ms

F:\Project Euler Solutions>

Code

// Project Euler 1-25
public class ProjectEulerSolutions {

	public static void main(String...args) {
        long startTime, endTime;

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 1:: "+problem1(1000));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 2:: "+problem2(4000000));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 3:: "+problem3(600851475143L));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 4:: "+problem4(1000));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 5:: "+problem5());
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 6:: "+problem6(100));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 7:: "+problem7(10001));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 8:: "+problem8(theNumber));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 9:: "+problem9(1000));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 10:: "+problem10(2000000));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 11:: "+problem11());
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 12:: "+problem12());
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 13:: "+problem13());
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 14:: "+problem14(1000000));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 15:: "+problem15(20));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 16:: "+problem16(1000));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 17:: "+problem17(1000));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 18:: "+problem18());
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 19:: "+problem19(1901,2001));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 20:: "+problem20(100));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 21:: "+problem21(10000));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 22:: "+problem22("names.txt"));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 23:: "+problem23(limit));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 24:: "+problem24(1000000));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");

        startTime = System.nanoTime();
        System.out.println("\nAnswer to problem 25:: "+problem25(1000000));
        endTime = System.nanoTime();
        System.out.println("in less than "+(double)(endTime-startTime)/1000000+" ms");
	}

    // The Fibonacci sequence is defined by the recurrence relation:
    // F(n) = F(nגˆ’1) + F(nגˆ’2), where F(1) = 1 and F(2) = 1.
    // Hence the first 12 terms will be:
    // F1 = 1
    // F2 = 1
    // F3 = 2
    // F4 = 3
    // F5 = 5
    // F6 = 8
    // F7 = 13
    // F8 = 21
    // F9 = 34
    // F10 = 55
    // F11 = 89
    // F12 = 144
    // The 12th term, F12, is the first term to contain three digits.
    // What is the index of the first term in the Fibonacci sequence to contain 1000 digits?
    private static String problem25(int num) {
        int i = 0;
        int cnt = 2;
        java.math.BigInteger limit = (new java.math.BigInteger("10")).pow(999);
        java.math.BigInteger[] fib = new java.math.BigInteger[3];

        fib[0] = java.math.BigInteger.ONE;
        fib[2] = java.math.BigInteger.ONE;

        while ((fib[i]).compareTo(limit) < 0) {
            i = (i + 1) % 3;
            cnt++;
            fib[i] = fib[(i + 1) % 3].add(fib[(i + 2) % 3]);
        }
        return Integer.toString(cnt);
    }

    // A permutation is an ordered arrangement of objects.
    // For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4.
    // If all of the permutations are listed numerically or alphabetically, we call it lexicographic order.
    // The lexicographic permutations of 0, 1 and 2 are:
    // 012   021   102   120   201   210
    // What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
    private static int[] perm = new int[] { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    private static String problem24(int num) {
        int count = 1;

        while (count < num) {
            int N = perm.length;
            int i = N-1;

            while (perm[i - 1] >= perm[i]) {
                i = i - 1;
            }

            int j = N;
            while (perm[j - 1] <= perm[i - 1]) {
                j = j - 1;
            }
            swap(i - 1, j - 1);

            i++;
            j = N;

            while (i < j) {
                swap(i - 1, j - 1);
                i++;
                j--;
            }
            count++;
        }

        String permNum = "";
        for (int k = 0; k < perm.length; k++) {
            permNum = permNum + perm[k];
        }
        return permNum;
    }

    private static void swap(int i, int j) {
        int k = perm[i];
        perm[i] = perm[j];
        perm[j] = k;
    }


    // A perfect number is a number for which the sum of its proper divisors is exactly equal to the number
    // For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
    // A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.
    // As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24.
    // By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers.
    // However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number
    // that cannot be expressed as the sum of two abundant numbers is less than this limit.
    // Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
    private static final int limit = 28123;
    private static boolean[] isAbundant = new boolean[limit + 1];
    private static String problem23(int limit) {
        for (int i = 1; i < isAbundant.length; i++)
            isAbundant[i] = isAbundant(i);
        
        int sum = 0;
        for (int i = 1; i <= limit; i++) {
            if (!isSumOf2Abundants(i))
                sum += i;
        }
        return Integer.toString(sum);
    }

    private static boolean isSumOf2Abundants(int num) {
        for (int i = 0; i <= num; i++) {
            if (isAbundant[i] && isAbundant[num - i]) {
                return true;
            }
        }
        return false;
    }

    private static boolean isAbundant(int num) {
        int sum = 0;
        for (int i=1; i<num; i++) {
            if (num%i==0) {
                sum += i;
            }
        }
        if (sum>num) {
            return true;
        } else {
            return false;
        }
    }

    // Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names,
    // begin by sorting it into alphabetical order. Then working out the alphabetical value for each name,
    // multiply this value by its alphabetical position in the list to obtain a name score.
    // For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53,
    // is the 938th name in the list. So, COLIN would obtain a score of 938 �— 53 = 49714.
    // What is the total of all the name scores in the file?
    private static String problem22(String filename) {
        int sum = 0;
        try {
            java.util.Scanner in = new java.util.Scanner(new java.io.File(filename)).useDelimiter("[\",]+");
            java.util.ArrayList<String> text = new java.util.ArrayList<String>();
            while (in.hasNext()) {
                text.add(in.next());
            }

            String[] names = new String[text.size()];
            for(int j=0; j<names.length; j++) {
                names[j] = text.get(j).toString();
            }

            java.util.Arrays.sort(names);

            for (int i = 0; i < names.length; i++) {
                int value = 0;
                for (int j = 0; j < names[i].length(); j++)
                    value += names[i].charAt(j) - 'A' + 1;
                sum += value * (i + 1);
            }
        } catch (Exception ex) {
            System.out.println("Something went wrong.");
        }
        return Integer.toString(sum);
    }

    private static String readFile(String filename) {
        java.io.File f = new java.io.File(filename);
        java.io.BufferedReader reader;
        String list = "";
        try {
            StringBuffer contents = new StringBuffer();
            String text = null;
            reader = new java.io.BufferedReader(new java.io.FileReader(f));
            while ((text = reader.readLine()) != null) {
                contents.append(text).append(System.getProperty("line.separator"));
            }
            list = contents.toString();
        } catch (java.io.FileNotFoundException e) {
            e.printStackTrace();
        } catch (java.io.IOException e) {
            e.printStackTrace();
        }
        return list;
    }

    // Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
    // If d(a) = b and d(b) = a, where a ג‰  b, then a and b are an amicable pair and each of a and b are called amicable numbers.
    // For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284.
    // The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
    // Evaluate the sum of all the amicable numbers under 10000.
    private static String problem21(int last) {
        int sum = 0;
        for (int i = 1; i < last; i++) {
            if (isAmicable(i))
                sum += i;
        }
        return Integer.toString(sum);
    }
    
    
    private static boolean isAmicable(int n) {
        int m = divisorSum(n);
        return m != n && divisorSum(m) == n;
    }
    
    private static int divisorSum(int n) {
        int sum = 0;
        for (int i = 1; i < n; i++) {
            if (n % i == 0)
                sum += i;
        }
        return sum;
    }

    // n! means n �— (n גˆ’ 1) �— ... �— 3 �— 2 �— 1
    // For example, 10! = 10 �— 9 �— ... �— 3 �— 2 �— 1 = 3628800,
    // and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
    // Find the sum of the digits in the number 100!
    private static String problem20(int last) {
        java.math.BigInteger prod = java.math.BigInteger.ONE;
        for (int i = 2; i <= last; i++) {
            prod = prod.multiply(java.math.BigInteger.valueOf(i));
        }
        String temp = prod.toString();
        int sum = 0;
        for (int i = 0; i < temp.length(); i++) {
            sum += temp.charAt(i) - '0';
        }
        return Integer.toString(sum);
    }

    // You are given the following information, but you may prefer to do some research for yourself.
    // 1 Jan 1900 was a Monday.
    // Thirty days has September,
    // April, June and November.
    // All the rest have thirty-one,
    // Saving February alone,
    // Which has twenty-eight, rain or shine.
    // And on leap years, twenty-nine.
    // A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
    // How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?
    private static String problem19(int year, int till) {
        int firstDay = 3; // 1 Jan 1901 was Tuesday 
        int sundays = 0;
        for(int i=year; i<till; i++) {
            int days_In_Month = getDays(1,i);
            int dif = days_In_Month%7;
            if(i!=year) {
                firstDay = (dif+firstDay)%7;
            }
            if(firstDay == 1) {
                sundays++;
            }
            for(int m=2; m<=12; m++) {
                firstDay = (dif+firstDay)%7;
                if(firstDay == 1) {
                    sundays++;
                }
                days_In_Month = getDays(m,i);
                dif = days_In_Month%7;
            }
        }
        return Integer.toString(sundays);
    }

    private static int getDays(int month, int year) {
        switch (month) {
            case 4:case 6: case 9: case 11:
                return 30;
            case 2:
                return (isLeapYear(year))?29:28;
        }
        return 31;
    }

    static boolean isLeapYear(int year) {
        boolean leap = false;
        if(year%4==0) {
            if(year%100==0) {
                leap = (year%400==0)?true:false;
            } else {
                leap = true;
            }
        }
        return leap;
    }

    // By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
    // 3
    // 7 4
    // 2 4 6
    // 8 5 9 3
    // That is, 3 + 7 + 4 + 9 = 23.
    // Find the maximum total from top to bottom of the triangle below:
    // 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
    // NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route.
    // However, Problem 67, is the same challenge with a triangle containing one-hundred rows;
    //  it cannot be solved by brute force, and requires a clever method! ;o)
    private static int[][] triangle = {
        {75},
        {95,64},
        {17,47,82},
        {18,35,87,10},
        {20, 4,82,47,65},
        {19, 1,23,75, 3,34},
        {88, 2,77,73, 7,63,67},
        {99,65, 4,28, 6,16,70,92},
        {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, 4,68,89,53,67,30,73,16,69,87,40,31},
        { 4,62,98,27,23, 9,70,98,73,93,38,53,60, 4,23},
    };
    private static String problem18() {
        for (int i = triangle.length - 2; i >= 0; i--) {
            for (int j = 0; j < triangle[i].length; j++) {
                triangle[i][j] += Math.max(triangle[i + 1][j], triangle[i + 1][j + 1]);
            }
        }
        return Integer.toString(triangle[0][0]);
    }

    // If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
    // If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?
    // NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 
    // 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.
    private static String[] under_twenty = new String[]{"", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};
    private static String[] tens = new String[]{"", "ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
    private static String hundred = "hundred";
    private static String thousand = "thousand";
    private static String and = "and";

    private static String problem17(int limit) {
        int sum = 0;
        for (int i = 1; i <= limit; i++) {
            sum += countLetters(i);
        }
        return Integer.toString(sum);
    }

    private static int countLetters(int num) {
        int letters_count = 0;
        
        if(num%100<20) {
            letters_count += under_twenty[num%100].length();
        } else {
            letters_count += under_twenty[num%10].length();
            letters_count += tens[(num%100-num%10)/10].length();
        }
        
        if(num>=100 && num<1000) {
            letters_count += under_twenty[(num%1000-num%100)/100].length();
            letters_count += hundred.length();
            if(num%100!=0) {
                letters_count+=and.length();
            }
        }
        
        if(num==1000) {
            letters_count += under_twenty[1].length();
            letters_count += thousand.length();
        }
        return letters_count;
    }

    // 2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
    // What is the sum of the digits of the number 2^1000?
    private static String problem16(int limit) {
        java.math.BigInteger n = java.math.BigInteger.valueOf(2);
        n = n.pow(limit);
        return Integer.toString(sumOfDigits(n.toString()));
    }

    private static int sumOfDigits(String s) {
        int sum = 0;

        for (int i = 0; i < s.length(); i++) {
            Character c = new Character(s.charAt(i));
            String z = c.toString();
            int j = Integer.parseInt(z);
            sum += j;
        }
        
        return sum;
    }

    // Starting in the top left corner of a 2�—2 grid, and only being able to move to the right and down,
    // there are exactly 6 routes to the bottom right corner.
    // How many such routes are there through a 20�—20 grid?
    private static String problem15(int size) {
        long paths = 1;
        for (int i = 0; i < size; i++) {
            paths *= (2 * size) - i;
            paths /= i + 1;
        }
        return Long.toString(paths);
    }

    // The following iterative sequence is defined for the set of positive integers:
    // n ג†’ n/2 (n is even)
    // n ג†’ 3n + 1 (n is odd)
    // Using the rule above and starting with 13, we generate the following sequence:
    // 13 ג†’ 40 ג†’ 20 ג†’ 10 ג†’ 5 ג†’ 16 ג†’ 8 ג†’ 4 ג†’ 2 ג†’ 1
    // It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms.
    // Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
    // Which starting number, under one million, produces the longest chain?
    // NOTE: Once the chain starts the terms are allowed to go above one million.
    private static String problem14(long number) {
        long maxLength = 1;
        int nr = -1;
        for (int i = 3; i < number; i++) {
            long sequenceSize = getSequenceSize(i);
            if (sequenceSize > maxLength) {
                maxLength = sequenceSize;
                nr = i;
            }
        }
        return Integer.toString(nr);
    }

    private static long getSequenceSize(int n) {
        long x = n;
        long count = 1;
        while (x > 1) {
            if ((x & 1) == 0) {        // x%2==0
                x = x >> 1;            // x=x/2
                count++;               // count=count+1
            } else {
                x = x + x + x + 1;     // x=3*x+1
                x = x >> 1;            // x=x/2
                count+=2;               
            }
        }
        return count;
    }


    // Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
    /* 
    37107287533902102798797998220837590246510135740250
    46376937677490009712648124896970078050417018260538
    74324986199524741059474233309513058123726617309629
    91942213363574161572522430563301811072406154908250
    23067588207539346171171980310421047513778063246676
    89261670696623633820136378418383684178734361726757
    28112879812849979408065481931592621691275889832738
    44274228917432520321923589422876796487670272189318
    47451445736001306439091167216856844588711603153276
    70386486105843025439939619828917593665686757934951
    62176457141856560629502157223196586755079324193331
    64906352462741904929101432445813822663347944758178
    92575867718337217661963751590579239728245598838407
    58203565325359399008402633568948830189458628227828
    80181199384826282014278194139940567587151170094390
    35398664372827112653829987240784473053190104293586
    86515506006295864861532075273371959191420517255829
    71693888707715466499115593487603532921714970056938
    54370070576826684624621495650076471787294438377604
    53282654108756828443191190634694037855217779295145
    36123272525000296071075082563815656710885258350721
    45876576172410976447339110607218265236877223636045
    17423706905851860660448207621209813287860733969412
    81142660418086830619328460811191061556940512689692
    51934325451728388641918047049293215058642563049483
    62467221648435076201727918039944693004732956340691
    15732444386908125794514089057706229429197107928209
    55037687525678773091862540744969844508330393682126
    18336384825330154686196124348767681297534375946515
    80386287592878490201521685554828717201219257766954
    78182833757993103614740356856449095527097864797581
    16726320100436897842553539920931837441497806860984
    48403098129077791799088218795327364475675590848030
    87086987551392711854517078544161852424320693150332
    59959406895756536782107074926966537676326235447210
    69793950679652694742597709739166693763042633987085
    41052684708299085211399427365734116182760315001271
    65378607361501080857009149939512557028198746004375
    35829035317434717326932123578154982629742552737307
    94953759765105305946966067683156574377167401875275
    88902802571733229619176668713819931811048770190271
    25267680276078003013678680992525463401061632866526
    36270218540497705585629946580636237993140746255962
    24074486908231174977792365466257246923322810917141
    91430288197103288597806669760892938638285025333403
    34413065578016127815921815005561868836468420090470
    23053081172816430487623791969842487255036638784583
    11487696932154902810424020138335124462181441773470
    63783299490636259666498587618221225225512486764533
    67720186971698544312419572409913959008952310058822
    95548255300263520781532296796249481641953868218774
    76085327132285723110424803456124867697064507995236
    37774242535411291684276865538926205024910326572967
    23701913275725675285653248258265463092207058596522
    29798860272258331913126375147341994889534765745501
    18495701454879288984856827726077713721403798879715
    38298203783031473527721580348144513491373226651381
    34829543829199918180278916522431027392251122869539
    40957953066405232632538044100059654939159879593635
    29746152185502371307642255121183693803580388584903
    41698116222072977186158236678424689157993532961922
    62467957194401269043877107275048102390895523597457
    23189706772547915061505504953922979530901129967519
    86188088225875314529584099251203829009407770775672
    11306739708304724483816533873502340845647058077308
    82959174767140363198008187129011875491310547126581
    97623331044818386269515456334926366572897563400500
    42846280183517070527831839425882145521227251250327
    55121603546981200581762165212827652751691296897789
    32238195734329339946437501907836945765883352399886
    75506164965184775180738168837861091527357929701337
    62177842752192623401942399639168044983993173312731
    32924185707147349566916674687634660915035914677504
    99518671430235219628894890102423325116913619626622
    73267460800591547471830798392868535206946944540724
    76841822524674417161514036427982273348055556214818
    97142617910342598647204516893989422179826088076852
    87783646182799346313767754307809363333018982642090
    10848802521674670883215120185883543223812876952786
    71329612474782464538636993009049310363619763878039
    62184073572399794223406235393808339651327408011116
    66627891981488087797941876876144230030984490851411
    60661826293682836764744779239180335110989069790714
    85786944089552990653640447425576083659976645795096
    66024396409905389607120198219976047599490197230297
    64913982680032973156037120041377903785566085089252
    16730939319872750275468906903707539413042652315011
    94809377245048795150954100921645863754710598436791
    78639167021187492431995700641917969777599028300699
    15368713711936614952811305876380278410754449733078
    40789923115535562561142322423255033685442488917353
    44889911501440648020369068063960672322193204149535
    41503128880339536053299340368006977710650566631954
    81234880673210146739058568557934581403627822703280
    82616570773948327592232845941706525094512325230608
    22918802058777319719839450180888072429661980811197
    77158542502016545090413245809786882778948721859617
    72107838435069186155435662884062257473692284509516
    20849603980134001723930671666823555245252804609722
    53503534226472524250874054075591789781264330331690
    */
    private static String problem13() {
        String numbers[] = { 
        "37107287533902102798797998220837590246510135740250",
        "46376937677490009712648124896970078050417018260538",
        "74324986199524741059474233309513058123726617309629",
        "91942213363574161572522430563301811072406154908250",
        "23067588207539346171171980310421047513778063246676",
        "89261670696623633820136378418383684178734361726757",
        "28112879812849979408065481931592621691275889832738",
        "44274228917432520321923589422876796487670272189318",
        "47451445736001306439091167216856844588711603153276",
        "70386486105843025439939619828917593665686757934951",
        "62176457141856560629502157223196586755079324193331",
        "64906352462741904929101432445813822663347944758178",
        "92575867718337217661963751590579239728245598838407",
        "58203565325359399008402633568948830189458628227828",
        "80181199384826282014278194139940567587151170094390",
        "35398664372827112653829987240784473053190104293586",
        "86515506006295864861532075273371959191420517255829",
        "71693888707715466499115593487603532921714970056938",
        "54370070576826684624621495650076471787294438377604",
        "53282654108756828443191190634694037855217779295145",
        "36123272525000296071075082563815656710885258350721",
        "45876576172410976447339110607218265236877223636045",
        "17423706905851860660448207621209813287860733969412",
        "81142660418086830619328460811191061556940512689692",
        "51934325451728388641918047049293215058642563049483",
        "62467221648435076201727918039944693004732956340691",
        "15732444386908125794514089057706229429197107928209",
        "55037687525678773091862540744969844508330393682126",
        "18336384825330154686196124348767681297534375946515",
        "80386287592878490201521685554828717201219257766954",
        "78182833757993103614740356856449095527097864797581",
        "16726320100436897842553539920931837441497806860984",
        "48403098129077791799088218795327364475675590848030",
        "87086987551392711854517078544161852424320693150332",
        "59959406895756536782107074926966537676326235447210",
        "69793950679652694742597709739166693763042633987085",
        "41052684708299085211399427365734116182760315001271",
        "65378607361501080857009149939512557028198746004375",
        "35829035317434717326932123578154982629742552737307",
        "94953759765105305946966067683156574377167401875275",
        "88902802571733229619176668713819931811048770190271",
        "25267680276078003013678680992525463401061632866526",
        "36270218540497705585629946580636237993140746255962",
        "24074486908231174977792365466257246923322810917141",
        "91430288197103288597806669760892938638285025333403",
        "34413065578016127815921815005561868836468420090470",
        "23053081172816430487623791969842487255036638784583",
        "11487696932154902810424020138335124462181441773470",
        "63783299490636259666498587618221225225512486764533",
        "67720186971698544312419572409913959008952310058822",
        "95548255300263520781532296796249481641953868218774",
        "76085327132285723110424803456124867697064507995236",
        "37774242535411291684276865538926205024910326572967",
        "23701913275725675285653248258265463092207058596522",
        "29798860272258331913126375147341994889534765745501",
        "18495701454879288984856827726077713721403798879715",
        "38298203783031473527721580348144513491373226651381",
        "34829543829199918180278916522431027392251122869539",
        "40957953066405232632538044100059654939159879593635",
        "29746152185502371307642255121183693803580388584903",
        "41698116222072977186158236678424689157993532961922",
        "62467957194401269043877107275048102390895523597457",
        "23189706772547915061505504953922979530901129967519",
        "86188088225875314529584099251203829009407770775672",
        "11306739708304724483816533873502340845647058077308",
        "82959174767140363198008187129011875491310547126581",
        "97623331044818386269515456334926366572897563400500",
        "42846280183517070527831839425882145521227251250327",
        "55121603546981200581762165212827652751691296897789",
        "32238195734329339946437501907836945765883352399886",
        "75506164965184775180738168837861091527357929701337",
        "62177842752192623401942399639168044983993173312731",
        "32924185707147349566916674687634660915035914677504",
        "99518671430235219628894890102423325116913619626622",
        "73267460800591547471830798392868535206946944540724",
        "76841822524674417161514036427982273348055556214818",
        "97142617910342598647204516893989422179826088076852",
        "87783646182799346313767754307809363333018982642090",
        "10848802521674670883215120185883543223812876952786",
        "71329612474782464538636993009049310363619763878039",
        "62184073572399794223406235393808339651327408011116",
        "66627891981488087797941876876144230030984490851411",
        "60661826293682836764744779239180335110989069790714",
        "85786944089552990653640447425576083659976645795096",
        "66024396409905389607120198219976047599490197230297",
        "64913982680032973156037120041377903785566085089252",
        "16730939319872750275468906903707539413042652315011",
        "94809377245048795150954100921645863754710598436791",
        "78639167021187492431995700641917969777599028300699",
        "15368713711936614952811305876380278410754449733078",
        "40789923115535562561142322423255033685442488917353",
        "44889911501440648020369068063960672322193204149535",
        "41503128880339536053299340368006977710650566631954",
        "81234880673210146739058568557934581403627822703280",
        "82616570773948327592232845941706525094512325230608",
        "22918802058777319719839450180888072429661980811197",
        "77158542502016545090413245809786882778948721859617",
        "72107838435069186155435662884062257473692284509516",
        "20849603980134001723930671666823555245252804609722",
        "53503534226472524250874054075591789781264330331690" };
        
        java.math.BigInteger sum = java.math.BigInteger.ZERO;
        for (int i = 0; i < numbers.length; i++) {
            java.math.BigInteger b = new java.math.BigInteger(numbers[i]);
            sum = sum.add(b);
        }
        return sum.toString().substring(0, 10);
    }

    // The sequence of triangle numbers is generated by adding the natural numbers.
    // So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
    // 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
    // Let us list the factors of the first seven triangle numbers:
    // 1: 1
    // 3: 1,3
    // 6: 1,2,3,6
    // 10: 1,2,5,10
    // 15: 1,3,5,15
    // 21: 1,3,7,21
    // 28: 1,2,4,7,14,28
    // We can see that 28 is the first triangle number to have over five divisors.
    // What is the value of the first triangle number to have over five hundred divisors?
    private static String problem12() {
        int iter = 1;
        int size = 1;
        int numFactors = 0;
        
        while (numFactors <= 500) {
            numFactors = countFactors(size);
            iter++;
            size += iter;
        }
        return Integer.toString(size-iter);
    }

    private static int countFactors(int f) {
        int factors = 0;
        for (int i = 1; i <= Math.sqrt(f); i++) {
            if (f % i == 0) {
                factors += 2;
            }
        }
        return factors;
    }

    // In the 20�—20 grid below, four numbers along a diagonal line have been marked in red.
    /* 
    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
    */
    // The product of these numbers is 26 �— 63 �— 78 �— 14 = 1788696.
    // What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20�—20 grid?
    private static String problem11() {
        int product = 0, largest = 0;

        // horizontal
        for(int i = 0; i < 20; i++) {
            for(int j = 0; j < 17; j++) {
                product = grid[i][j] * grid[i][j+1] * grid[i][j+2] * grid[i][j+3];
                if(product > largest) {
                    largest = product;
                }
            }   
        }

        // vertical
        for(int i = 0; i < 17; i ++) {
            for(int j = 0; j < 20; j++) {
                product = grid[i][j] * grid[i+1][j] * grid[i+2][j] * grid[i+3][j];
                if(product > largest) {
                    largest = product;
                }
            }
        }

        // diagonal right
        for(int i = 0; i < 17; i++) {
            for(int j = 0; j < 17; j++) {
                product = grid[i][j] * grid[i+1][j+1] * grid[i+2][j+2] * grid[i+3][i+3];
                if(product > largest) {
                    largest = product;
                }
            }
        }

        // diagonal left
        for(int i = 0; i < 17; i ++) {
            for(int j = 3; j < 20; j ++) {
                product = grid[i][j] * grid[i+1][j-1] * grid[i+2][j-2] * grid[i+3][j-3];
                if(product > largest) {
                    largest = product;
                }
            }
        }
        return Integer.toString(largest);
    }

    private static int grid [][] = {
        {8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,8},
        {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},
        {67,26,20,68,02,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21},
        {24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
        {21,36,23,9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95},
        {78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,9,53,56,92},
        {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},
        {04,52,8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66},
        {88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69},
        {04,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36},
        {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}
        };

    // The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
    // Find the sum of all the primes below two million.
    private static String problem10(long upperLimit) {
        long count = 0;
        for (long i=2; i<2000000; i++) {
            if (SieveIsPrime(i)) {
                count += i;
            }
        }
        return Long.toString(count);
    }

    private static java.util.List<Long> listOfPrimes = new java.util.ArrayList<Long>();
    private static boolean SieveIsPrime(long n) {
        String strFromN = new Long(n).toString();
        if ((strFromN.length() != 1) && (strFromN.endsWith("2") || strFromN.endsWith("4") || strFromN.endsWith("5") || strFromN.endsWith("6") || strFromN.endsWith("8"))) {
            return false;
        }

        for (Long num : listOfPrimes) {
            if (num > Math.sqrt(n)) {
                break;
            }
            if (n % num.longValue() == 0) {
                return false;
            }
        }


        listOfPrimes.add(new Long(n));
        return true;
    }

    // A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
    // a^2 + b^2 = c^2
    // For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
    // There exists exactly one Pythagorean triplet for which a + b + c = 1000.
    // Find the product abc.
    private static String problem9(int theNumber) {
        int a = 0, b = 0, c = 0;
        int sum = theNumber;
        boolean flag = false;
        for(a=1; a<sum/3; a++) {
            for(b=a; b<sum/2; b++) {
                c = sum - a - b;
                if(a*a + b*b == c*c) {
                    flag = true;
                    break;
                }
            }
            if(flag) {
                break;
            }
        }
        return Integer.toString(a*b*c);
    }

    // The four adjacent digits in the 1000-digit number that have the greatest product are 9 �— 9 �— 8 �— 9 = 5832.
    // Find the thirteen adjacent digits in the 1000-digit number that have the greatest product.
    // What is the value of this product?
    private static String problem8(String theNumber) {
        long maxProd = -1;
        for (int i = 0; i + ADJACENT <= theNumber.length(); i++) {
            long prod = 1;
            for (int j = 0; j < ADJACENT; j++) {
                prod *= theNumber.charAt(i + j) - '0';
            }
            maxProd = Math.max(prod, maxProd);
        }
        return Long.toString(maxProd);
    }

    private static final int ADJACENT = 13;
    private static final String theNumber = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";

    // By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
    // What is the 10001st prime number?
    private static String problem7(long limit) {
        for (int i = 2, count = 0; ; i++) {
            if (isPrime(i)) {
                count++;
                if (count == limit)
                    return Integer.toString(i);
            }
        }
    }

    private static boolean isPrime(long number) {
        if(number<=1) {
            return false;
        }
        if(number==2) {
            return true;
        }
        if(number%2==0) {
            return false;
        }
        for(long i=3; i<java.lang.Math.sqrt(number)+1; i++) {
            if(number%i==0) {
                return false;
            }
        }
        return true;
    }

    // The sum of the squares of the first ten natural numbers is,
    // 12 + 22 + ... + 102 = 385
    // The square of the sum of the first ten natural numbers is,
    // (1 + 2 + ... + 10)2 = 552 = 3025
    // Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 גˆ’ 385 = 2640.
    // Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
    private static String problem6(long limit) {
        // The sum of natural numbers can be expressed as: n(n + 1) / 2 => 1^2 + 2^2 + ... + 10^2 = 385
        // The sum of squares of natural numbers can be expressed as: n(n + 1)(2n + 1) / 6 => (1 + 2 + ... + 10)^2 = 55^2 = 3025
        long sum = 0;
        long squared = 0;
        long result = 0;
        long n = limit;
        for (int i = 1; i <= n; i++) {
            sum = n * (n+1)/ 2;
            squared = (n * (n + 1) * (2 * n + 1)) / 6;
        }
        result = sum * sum - squared;
        return Long.toString(result);
    }

    // 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
    // What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
    private static String problem5() {
        // lcm(a,b) = (a*b)/gcd(a,b)
        // where: LCM = Least Common Multiple, GCD = Greatest Common Divisor
        long num = 1;
        for(int i=1; i<=20; i++) {
            num = lcm(num,i);
        }
        return Long.toString(num);
    }

    private static long gcd(long a, long b) {
        while (b != 0) {
            long tmp = b;
            b = a % b;
            a = tmp;
        }
        return a;
    }

    private static long lcm(long a, long b) {
        return (a*b) / gcd(a,b);
    }

    // A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 �— 99.
    // Find the largest palindrome made from the product of two 3-digit numbers.
    private static String problem4(int limit) {
        int maxPalindrome = -1;
        for (int i = 100; i < limit; i++) {
            for (int j = 100; j < limit; j++) {
                int product = i * j;
                StringBuilder productString = new StringBuilder(Integer.toString(product));
                if(productString.toString().equals(productString.reverse().toString()) && product > maxPalindrome) {
                    maxPalindrome = product;
                }
            }
        }
        return Integer.toString(maxPalindrome);
    }

    // The prime factors of 13195 are 5, 7, 13 and 29.
    // What is the largest prime factor of the number 600851475143 ?
    private static String problem3(long num) {
        // Fundamental Theorem of Arithmetic: Any integer greater than 1 is either a prime number, 
        // or can be written as a unique product of prime numbers (ignoring the order).
        long newNum = num;
        long largestFact = 0;

        int counter = 2;
        while (counter * counter <= newNum) {
            if (newNum % counter == 0) {
                newNum = newNum / counter;
                largestFact = counter;
            } else {
                counter++;
            }
        }
        if (newNum > largestFact) {
            largestFact = newNum;
        }
        return Long.toString(largestFact);
    }

    // Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
    // 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
    // By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
    private static String problem2(int limit) {
        int prevTerm = 1;
        int nextTerm = 2;
        int sum = 0;
        while(prevTerm<=limit) {
            if(prevTerm%2==0) {
                sum += prevTerm;
            }
            int temp = prevTerm;
            prevTerm = nextTerm;
            nextTerm = temp + prevTerm;
        }
        return Integer.toString(sum);
    }

    // If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9.
    // The sum of these multiples is 23.
    // Find the sum of all the multiples of 3 or 5 below 1000.
    private static String problem1(int below) {
        int total = 0;
        for(int i=0; i<below; i++) {
            if(i%3==0 || i%5==0) {
                total += i;
            }
        }
        return Integer.toString(total);
    }
}