practice.js

/*global console, FileReader*/
/*jslint continue:true*/

/* Problem 1
 * 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. */
function multiples35(end) {
    "use strict";
    var i = 0,
        result = 0;
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 1000;
    }
    
    for (i = 0; i < end; i += 1) {
        if ((i % 3) === 0 || (i % 5) === 0) {
            result += i;
        }
    }
    return result;
}

/* Problem 2
 * 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. */
function evenFib(a, b, sum) {
    "use strict";
    var end = 4000000,
        current = a + b,
        result = 0;
    //console.log(current + ", " + sum);
    if (current > end) {
        return sum;
    } else if (current % 2 === 0) {
        result = evenFib(b, current, sum + current);
    } else {
        result = evenFib(b, current, sum);
    }
    return result;
}
function startEvenFib() {
    "use strict";
    var result = evenFib(1, 1, 0);
    return result;
}

/* Problem 3
 * The prime factors of 13195 are 5, 7, 13 and 29.
 * What is the largest prime factor of the number 600851475143 ? */
function largestPrimeFactor(n) {
    "use strict";
    var factors = [],
        i = 0;
    while (n % 2 === 0) {
        if (factors.indexOf(2) === -1) {
            factors.push(2);
        }
        n = n / 2;
    }
    for (i = 3; i < Math.sqrt(n); i += 2) {
        while (n % i === 0) {
            if (factors.indexOf(i) === -1) {
                factors.push(i);
            }
            n = n / i;
        }
    }
    if (n > factors[factors.length - 1]) {
        return n;
    }
    return factors.pop;
}

/* Problem 4
 * 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. */
function largestPalendrome(end) {
    "use strict";
    var i, j, k, product, palendromes = [];
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 999;
    }
    
    // start at end and iterate down
    for (i = end; i > 0; i -= 1) {
        for (j = end; j > 0; j -= 1) {
            // check for palendrome
            product = (i * j).toString();
            for (k = 0; k < product.length / 2; k += 1) {
                if (product[k] !== product[product.length - k - 1]) {
                    break;
                }
            }
            if (k === (product.length / 2)) {
                palendromes.push(parseInt(product, 10));
            }
        }
    }
    product = 0;
    for (i = 0; i < palendromes.length; i += 1) {
        if (palendromes[i] > product) {
            product = palendromes[i];
        }
    }
    return product;
}

/* Problem 5
 * 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? */
function smallestDividingNumber(end) {
    "use strict";
    var i, n;
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 20;
    }
    
    n = end;
    while (true) {
        for (i = end; i > end / 2; i -= 1) {
            if (n % i !== 0) {
                break;
            }
        }
        if (i === end / 2) {
            return n;
        }
        n += end;
    }
}

/* Problem 6
 * The sum of the squares of the first ten natural numbers is,
 *      1^2 + 2^2 + ... + 10^2 = 385
 * The square of the sum of the first ten natural numbers is,
 *      (1 + 2 + ... + 10)^2 = 55^2 = 3025
 * Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
 * Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum. */
function findSquaredSumDif(end) {
    "use strict";
    var i,
        sumSquared = 0,
        squaredSum = 0;
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 100;
    }
    
    for (i = 1; i <= end; i += 1) {
        sumSquared += i;
        squaredSum += (i * i);
    }
    sumSquared *= sumSquared;
    return sumSquared - squaredSum;
}

/* Problem 7
 * 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 10,001st prime number? */
function isPrime(n) {
    "use strict";
    var i;
    n = Math.abs(n);
    for (i = 2; i <= Math.sqrt(n); i += 1) {
        if (n % i === 0) {
            return false;
        }
    }
    return true;
}

function findXPrimeNumber(end) {
    "use strict";
    var n = 2,
        counter = 0;
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 10001;
    }
    
    while (true) {
        if (isPrime(n)) {
            counter += 1;
        }
        if (counter === end) {
            return n;
        }
        n += 1;
    }
}

/* Problem 8
 * 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? */
function maxAdjacentProduct(end) {
    "use strict";
    var i,
        flag = 0,
        result = 1,
        product = 1,
        number = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 13;
    }
    
    // set up initial number
    for (i = 0; i < end; i += 1) {
        product *= parseInt(number[i], 10);
    }
    result = product;
    // iterate through looking for windows without 0s finding max
    for (i = 0; i < number.length - end; i += 1) {
        if (number[i + end] === "0") {
            flag = end;
        }
        if (number[i] === "0" && number[i + end] !== "0") {
            product *= parseInt(number[i + end], 10);
        } else if (number[i + end] === "0" && number[i] !== "0") {
            product /= parseInt(number[i], 10);
        } else if (number[i + end] !== "0" && number[i] !== "0") {
            product /= parseInt(number[i], 10);
            product *= parseInt(number[i + end], 10);
            if (product > result && flag <= 0) {
                result = product;
            }
        }
        flag -= 1;
    }
    return result;
}

/* Problem 9
 * 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. */
function findPTriplet(a, b) {
    "use strict";
    var i, j, c, e;
    c = Math.sqrt((a * a) + (b * b));
    e = a + b + c;
    //console.log("a: " + a + ", b: " + b + ", c: " + c + ", e: " + e);
    if (e === 1000) {
        return a * b * c;
    }
    for (i = a + 1; i < 400; i += 1) {
        for (j = 2; j < 401; j += 1) {
            if (Math.sqrt((i * i) + (j * j)) % 1 === 0) {
                return findPTriplet(i, j);
            }
        }
    }
}

function startFindPTriplet() {
    "use strict";
    return findPTriplet(0, 1);
}

/* Problem 10
 * The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
 * Find the sum of all the primes below two million. */
function findXPrimeNumber(end) {
    "use strict";
    var n = 2,
        sum = 0;
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 2000000;
    }
    
    while (n < end) {
        if (isPrime(n)) {
            sum += n;
        }
        n += 1;
    }
    return sum;
}

/* Problem 11
 * In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
 * 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? */
function findGridMaxProduct() {
    "use strict";
    var i, j, k, max,
        grid = [[8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8],
                [49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0],
                [81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65],
                [52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 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, 3, 45, 2, 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, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21],
                [24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72],
                [21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95],
                [78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92],
                [16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57],
                [86, 56, 0, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58],
                [19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40],
                [4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66],
                [88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69],
                [4, 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, 4, 36, 16],
                [20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54],
                [1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48]
            ],
        lrProduct = 1, udProduct = 1, lrudProduct = 1, lrduProduct = 1, end = 4, result = 0;

    // go through y
    for (i = 0; i < grid.length; i += 1) {
        // go through x
        for (j = 0; j < grid[i].length; j += 1) {
            // go through length (4)
            lrProduct = udProduct = lrudProduct = lrduProduct = 1;
            for (k = 0; k < end; k += 1) {
                // left and right
                if (j < grid[i].length - end) {
                    lrProduct *= grid[i][j + k];
                }
                // up and down
                if (i < grid.length - end) {
                    udProduct *= grid[i + k][j];
                }
                // left -> right, up -> down
                if ((j < grid[i].length - end) && (i < grid.length - end)) {
                    lrudProduct *= grid[i + k][j + k];
                }
                // left -> right, down -> up
                if ((j < grid[i].length - end) && (i >= end - 1)) {
                    lrduProduct *= grid[i - k][j + k];
                }
            }
            //console.log("lr: " + lrProduct + ", ud: " + udProduct + ", lrud: " + lrudProduct + ", lrdu: " + lrduProduct);
            k = (lrProduct > udProduct) ? lrProduct : udProduct;
            max = (lrudProduct > lrduProduct) ? lrudProduct : lrduProduct;
            max = (max > k) ? max : k;
            result = (result > max) ? result : max;
        }
    }
    return result;
}

/* Problem 12
 * 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? */
function numDivisors(n) {
    "use strict";
    var i, result = 0;
    for (i = 1; i <= Math.sqrt(n); i += 1) {
        if (i === Math.sqrt(n)) {
            result += 1;
        } else if (n % i === 0) {
            result += 2;
        }
    }
    return result;
}
function findTriNumXDivisors(end) {
    "use strict";
    var i = 1,
        divisors = 0,
        triangleNumber = 0;
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 500;
    }
    
    while (divisors <= end) {
        triangleNumber += i;
        divisors = numDivisors(triangleNumber);
        i += 1;
        //console.log(triangleNumber + ": " + divisors);
    }
    return triangleNumber;
}

/* Problem 13
 * Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
 * see array below for numbers */

function firstTenSum() {
    "use strict";
    var i, sum = 0,
        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];
    for (i = 0; i < numbers.length; i += 1) {
        sum += numbers[i];
    }
    console.log(sum);
    sum = sum.toString();
    sum = sum.substring(0, 1) + sum.substring(2, 11);
    return sum;
}

/* Problem 14
 * The following iterative sequence is defined for the set of positive integers:
 *   n → n/2 (n is even)
 *   n → 3n + 1 (n is odd)
 * Using the rule above and starting with 13, we generate the following sequence:
 *        13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 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. */
function collatz(n, c) {
    "use strict";
    c += 1;
    if (n === 1) {
        return c;
    }
    if (n % 2 === 0) {
        return collatz(n / 2, c);
    } else {
        return collatz(3 * n + 1, c);
    }
}
function findLongestCSequence(end) {
    "use strict";
    var i, result, temp, max = 0;
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 1000000;
    }
    
    for (i = 1; i < end; i += 1) {
        temp = collatz(i, 0);
        if (temp > max) {
            max = temp;
            result = i;
        }
    }
    return result;
}

/* Problem 15
 * 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? */
function findEnd(x, y, max, count) {
    "use strict";
    if (x === max && y === max) {
        if (count % 1000000000 === 0) {
            console.log(count);
        }
        return count + 1;
    }
    if (x < max) {
        count = findEnd(x + 1, y, max, count);
    }
    if (y < max) {
        count = findEnd(x, y + 1, max, count);
    }
    return count;
}
function numGridPaths(n) {
    "use strict";
    // sanatize input
    n = parseInt(n, 10);
    if (isNaN(n)) {
        n = 20;
    }
    return findEnd(0, 0, n, 0);
}
/* alternate solution found online in python and translated
 * 1. S[i][j] = 1                      if j = 0
 * 2. S[i][j] = S[i][j-1] + S[i-1][j]  if 0 < j < i
 * 3. S[i][j] = 2 * S[i][j-1]          if i = j     */
function dynNumGridPaths(n) {
    "use strict";
    var i, j, result = [];
    
    // sanatize input
    n = parseInt(n, 10);
    if (isNaN(n)) {
        n = 20;
    }
    
    // 1. S[i][j] = 1  if j = 0
    for (i = 0; i < n; i += 1) {
        result[i] = 1;
    }
    
    for (i = 1; i <= n; i += 1) {
        // 2. S[i][j] = S[i][j-1] + S[i-1][j]  if 0 < j < i
        for (j = 1; j < i; j += 1) {
            result[j] = result[j - 1] + result[j];
        }
        // 3. S[i][j] = 2 * S[i][j-1]  if i = j
        result[i] = 2 * result[i - 1];
    }
    //console.log(result);
    return result[n];
}

/* Problem 16
 * 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? */
function sumTwoHighPower(end) {
    "use strict";
    var i, j, result = 0,
        carry = 0,
        exp = [1];
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 1000;
    }
    
    // iterate to exponent times
    for (i = 0; i < end; i += 1) {
        // iterate through array multiplying by 2
        for (j = 0; j < exp.length; j += 1) {
            exp[j] *= 2;
            exp[j] += carry;
            carry = 0;
            if (exp[j] >= 10) {
                exp[j] %= 10;
                carry = 1;
                if (exp[j + 1] === undefined) {
                    exp[j + 1] = 1;
                    carry = 0;
                    break;
                }
            }
        }
        //console.log(exp);
    }
    
    for (i = 0; i < exp.length; i += 1) {
        result += exp[i];
    }
    
    return result;
}

/* Problem 17
 * 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. */
function translateSum(end) {
    "use strict";
    // 21-30, 27-90, 28 00
    var num, i, j, result = 0,
        translate = [4, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8, 6, 6, 5, 5, 5, 7, 6, 6];
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 1000;
    }
    
    // iterate through all numbers <= 1000
    for (i = 1; i <= end; i += 1) {
        // thousands case
        if (i === 1000) {
            result += 11;
            break;
        }
        // hundreds case
        if (i >= 100) {
            num = i.toString();
            num = parseInt(num[0], 10);
            // number + hundred
            result += (translate[num] + 7);
            // check for "and"
            if (i % 100 !== 0) {
                result += 3;
            }
        }
        j = i % 100;
        if (j >= 20) {
            num = j.toString();
            num = parseInt(num[0], 10);
            result += translate[num + 18];
            j = j % 10;
        }
        if (j > 0) {
            result += translate[j];
        }
        //console.log(result);
    }
    return result;
}

/* Problem 18
 * 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) */

function longestPath() {
    "use strict";
    var i, j, a, b, 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]];
    
    for (i = triangle.length - 2; i >= 0; i -= 1) {
        for (j = 0; j < triangle[i].length; j += 1) {
            a = triangle[i][j] + triangle[i + 1][j];
            b = triangle[i][j] + triangle[i + 1][j + 1];
            triangle[i][j] = (a > b) ? a : b;
        }
    }
    // console.log(triangle);
    return triangle[0][0];
}

/* Problem 19
 * 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)? */
// NOTES: 364 Days in 52 weeks, 365 in year, 366 in leap year
function Sundaysin20thCentury(end) {
    "use strict";
    var i, j, result = 0, months = [31 % 7, 28 % 7, 31 % 7, 30 % 7, 31 % 7, 30 % 7, 31 % 7, 31 % 7, 30 % 7, 31 % 7, 30 % 7, 31 % 7],
        // mondays are 0, sundays are 6
        // 1900 starts at 0, so 1901 is 0 + 29 % 7 = 1
        start = 1;
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 2000;
    }
    
    for (i = 1901; i <= end; i += 1) {
        for (j = 0; j < months.length; j += 1) {
            if (start === 6) {
                result += 1;
            }
            start = (start + months[j]) % 7;
            if (j === 1 && i % 4 === 0 && (i % 100 !== 0 || i % 400 === 0)) {
                start = (start + 1) % 7;
            }
        }
    }
    
    return result;
}

/* Problem 20
 * n! means n × (n − 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! */
function factorialResultSum() {
    "use strict";
    var i, j, k, b, c, result = 0, product, t = [];
    
    product = [0, 0, 1];
    for (i = 99; i > 0; i -= 1) {
        b = [];
        t[0] = [];
        t[1] = [0];
        // set current value
        b.push(i % 10);
        if (i >= 10) {
            b.push(Math.floor(i / 10));
        }
        // iterate through and multiply to get 2 temps
        
        for (j = 0; j < b.length; j += 1) {
            c = 0;
            for (k = 0; k < product.length; k += 1) {
                t[2] = b[j] * product[k];
                t[j].push((t[2] + c) % 10);
                c = Math.floor((t[2] + c) / 10);
            }
            if (c !== 0) {
                t[j].push(c);
            }
        }
        // iterate though temps and add to get product
        c = 0;
        product = [];
        k = (t[0].length > t[1].length) ? t[0].length : t[1].length;
        for (j = 0; j < k; j += 1) {
            if (!t[0][j]) {
                t[0][j] = 0;
            }
            if (!t[1][j]) {
                t[1][j] = 0;
            }
            product.push((t[0][j] + t[1][j] + c) % 10);
            c = Math.floor((t[0][j] + t[1][j] + c) / 10);
        }
        if (c !== 0) {
            product.push(c);
        }
        //console.log(product);
    }
    // find the sum of values in product array
    for (i = 0; i < product.length; i += 1) {
        result += product[i];
    }
    return result;
}

/* Problem 21
 * 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 != 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.*/
function findDivisors(n) {
    "use strict";
    var i, result = [1];
    for (i = 2; i <= Math.sqrt(n); i += 1) {
        if (i === Math.sqrt(n)) {
            result.push(i);
        } else if (n % i === 0) {
            result.push(i);
            result.push(n / i);
        }
    }
    return result;
}

function amicableNumbers() {
    "use strict";
    var i, j, d, result = 0, aN1, aN2, aN = [];
    
    for (i = 0; i < 10000; i += 1) {
        d = findDivisors(i);
        aN1 = 0;
        for (j = 0; j < d.length; j += 1) {
            aN1 += d[j];
        }
        //console.log(aN1);
        d = findDivisors(aN1);
        aN2 = 0;
        for (j = 0; j < d.length; j += 1) {
            aN2 += d[j];
        }
        //console.log(aN2);
        if (aN.indexOf(aN2) === -1 && aN1 !== aN2 && aN2 === i) {
            aN.push(aN1);
            aN.push(aN2);
        }
    }
    // console.log(aN);
    for (i = 0; i < aN.length; i += 1) {
        result += aN[i];
    }
    return result;
}

/* Problem 22
 * Using names.txt, 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? */
function nameScore(name, pos) {
    "use strict";
    var letter, score = 0, rank = [0, "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"];

    for (letter = 0; letter < name.length; letter += 1) {
        score += rank.indexOf(name[letter]);
    }
    //console.log(score);
    score *= pos;
    return score;
}

document.getElementById('p22file').onchange = function () {
    "use strict";
    var i, file = this.files[0], reader = new FileReader();
    reader.onload = function (progressEvent) {
        var result = 0, word, words = this.result.replace(/['"]+/g, '').split(',');
        words.sort();
        
        for (word = 0; word < words.length; word += 1) {
            result += nameScore(words[word], word + 1);
            if (words[word] === "COLIN") {
                console.log(words[word], word + 1, nameScore(words[word], word + 1));
            }
            //console.log(result);
        }
        console.log(result);
        return result;
    };
    reader.readAsText(file);
};

/* Problem 23
 * 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. */
function nonAbundantSums() {
    "use strict";
    var i, temp, n, result, aNumbers = [], numbers = [], nonSums = [];
    // create a pool of abundant numbers
    for (n = 12; n <= 28111; n += 1) {
        // find the sum of divisors
        temp = 1;
        for (i = 2; i <= Math.sqrt(n); i += 1) {
            if (i === Math.sqrt(n)) {
                temp += i;

            } else if (n % i === 0) {
                temp += i;
                temp += (n / i);
            }
        }
        // add abundant number to array
        if (temp > n) {
            aNumbers.push(n);
        }
    }
    
    // create a list of all numbers
    for (n = 0; n <= 28123; n += 1) {
        numbers[n] = n;
    }
    
    // iterate through all numbers, removing sums of abundant numbers
    for (n = 0; n < aNumbers.length; n += 1) {
        for (i = 0; i < aNumbers.length; i += 1) {
            temp = aNumbers[n] + aNumbers[i];
            if (temp > 28123) {
                break;
            }
            numbers[temp] = 0;
        }
    }
    
    temp = 0;
    // iterate through whats left and get the sum
    for (i = 0; i < numbers.length; i += 1) {
        temp += numbers[i];
    }
    
    return temp;
}

/* Problem 24
 * 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? */
function lexicographicPermutations(n, d, e) {
    "use strict";
    // 3,628,800  362,880  40,320  5040  720  120  24  6  2  1
    var temp, i, counter;
        
    // check to make sure there are numbers left
    if (d.length < 1) {
        return n;
    }
    
    // find bracket of permutations
    temp = 1;
    for (i = 1; i < d.length; i += 1) {
        temp *= i;
    }
    
    // find # in bracket
    counter = 0;
    while (true) {
        if (e - temp > 0) {
            e -= temp;
            counter += 1;
        } else {
            break;
        }
    }
    
    // update number and digits and call function
    n = n * 10 + d[counter];
    d.splice(counter, 1);
    return lexicographicPermutations(n, d, e);
}
function findLexicographicPermutations() {
    "use strict";
    return lexicographicPermutations(0, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], 1000000);
}

/* Problem 25
 * The Fibonacci sequence is defined by the recurrence relation:
 * Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 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? */
function thousandDFib(a, b, i) {
    "use strict";
    var j, c = 0;
    
    // add 2 numbers as arrays
    for (j = 0; j < b.length; j += 1) {
        if (isNaN(a[j])) {
            a[j] = 0;
        }
        a[j] += (b[j] + c);
        if (a[j] >= 10) {
            a[j] -= 10;
            c = 1;
        } else {
            c = 0;
        }
    }
    if (c === 1) {
        a.splice(a.length, 0, 1);
    }
    
    // end condition, if length is 1000 digits
    if (a.length >= 1000) {
        console.log(a);
        return i;
    }
    
    // recursive call
    return thousandDFib(b, a, i + 1);
}
function startThousandDFib() {
    "use strict";
    var result = thousandDFib([1], [1], 3);
    return result;
}

/* Problem 26
 * A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
 *   1/2    =   0.5
 *   1/3    =   0.(3)
 *   1/4    =   0.25
 *   1/5    =   0.2
 *   1/6    =   0.1(6)
 *   1/7    =   0.(142857)
 *   1/8    =   0.125
 *   1/9    =   0.(1)
 *   1/10   =   0.1
 * Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
 * Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part. */
function recurringCycleFraction(end) {
    "use strict";
    var d, n, r, i, loop = true, result = [], maxN = 0, maxD = 0;
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 1000;
    }
    
    // iterate through numbers
    for (d = 2; d < end; d += 1) {
        n = 1;
        // repeat and divide
        while (loop) {
            if (n < d) {
                n *= 10;
            } else {
                r = n % d;
                // check if non-repeating
                if (r === 0) {
                    loop = false;
                    break;
                }
                // check if repeating restarts
                for (i = 0; i < result.length; i += 1) {
                    if (r === result[i]) {
                        loop = false;
                        break;
                    }
                }
                if (loop) {
                    result.push(r);
                    n = r;
                }
            }
        }
        loop = true;
        // check length of repeating num
        if (result.length > maxN) {
            maxN = result.length;
            maxD = d;
        }
        result = [];
    }
    return maxD;
}

/* Problem 27
 * Euler discovered the remarkable quadratic formula:
 *      n^2 + n + 41
 * It turns out that the formula will produce 40 primes for the consecutive integer values 0 ≤ n ≤ 39.
 * However, when n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41^2 + 41 + 41 is clearly divisible by 41.
 *
 * The incredible formula n^2 − 79n + 1601 was discovered, which produces 80 primes for the consecutive values 0 ≤ n ≤ 79.
 * The product of the coefficients, −79 and 1601, is −126479.
 *
 * Considering quadratics of the form:
 *      n^2 + an + b, where |a| < 1000 and |b| ≤ 1000
 *      where |n| is the modulus/absolute value of n
 *      e.g. |11| = 11 and |−4| = 4
 * Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0. */

/* function isPrime(n) {
    "use strict";
    var i;
    n = Math.abs(n);
    for (i = 2; i <= Math.sqrt(n); i += 1) {
        if (n % i === 0) {
            return false;
        }
    }
    return true;
} */
function consecutivePrimeGen(end) {
    "use strict";
    var n, a, b, temp, count, ans = {a: 0, b: 0, c: 0};
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 1000;
    }
    
    // find values for a, b, n in: n^2 + an + b
    for (a = 1 - end; a < end; a += 1) {
        for (b = -end; b <= end; b += 1) {
            // console.log(a, b, count);
            count = 0;
            for (n = 0; n < Math.abs(b); n += 1) {
                temp = (n * n) + (a * n) + b;
                //console.log(temp, isPrime(temp));
                if (isPrime(temp)) {
                    count += 1;
                } else {
                    break;
                }
            }
            if (count > ans.c) {
                ans.a = a;
                ans.b = b;
                ans.c = count;
                console.log(ans);
            }
        }
    }
    
    //console.log(ans);
    return ans.a * ans.b;
}

/* Problem 28
 * Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
 *    21 22 23 24 25
 *    20  7  8  9 10
 *    19  6  1  2 11
 *    18  5  4  3 12
 *    17 16 15 14 13
 * It can be verified that the sum of the numbers on the diagonals is 101.
 * What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way? */
function spiralSum(end) {
    "use strict";
    var i, c, n = 1, k = 2, sum = 1;
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 1001;
    }
    
    end -= 1;
    end /= 2;
    
    // go through all the diagnals
    for (i = 0; i < end; i += 1) {
        // get all 4 corners
        for (c = 1; c <= 4; c += 1) {
            n += k;
            sum += n;
        }
        k += 2;
    }
    return sum;
}

/* Problem 29  - DOES NOT WORK IN JS, SEE PHP
 * Consider all integer combinations of a^b for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
 *    2^2 = 4,  2^3 = 8,   2^4 = 16,  2^5 = 32
 *    3^2 = 9,  3^3 = 27,  3^4 = 81,  3^5 = 243
 *    4^2 = 16, 4^3 = 64,  4^4 = 256, 4^5 = 1024
 *    5^2 = 25, 5^3 = 125, 5^4 = 625, 5^5 = 3125
 * If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
 *    4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
 * How many distinct terms are in the sequence generated by a^b for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100? */
function distictPowerTerms(end) {
    "use strict";
    var a, b, res = [], out = [];
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 100;
    }
    
    for (a = 2; a <= end; a += 1) {
        for (b = 2; b <= end; b += 1) {
            res.push(Math.pow(a, b));
        }
    }

    res.sort(function (a, b) { return a - b; });
    if (res.length > 0) {
        for (a = 0; a < res.length - 1; a += 1) {
            if (res[a] !== res[a + 1]) {
                out.push(res[a]);
            }
        }
        out.push(res[res.length - 1]);
    }

    console.log(res);
    console.log(out);
    return out.length;
}


/* Problem 30
 * Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
 *      1634 = 1^4 + 6^4 + 3^4 + 4^4
 *      8208 = 8^4 + 2^4 + 0^4 + 8^4
 *      9474 = 9^4 + 4^4 + 7^4 + 4^4
 * As 1 = 1^4 is not a sum it is not included.
 * The sum of these numbers is 1634 + 8208 + 9474 = 19316.
 * Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.*/

/* This problem requires some pre-work. the highest number possible is 9^5 = 59049
 * the sum of x digits of these are as follows:
 *      n   total   digits
 *      1   59049   5   // this means we can have at least 5 digits
 *      5   295245  6   // this means we can have at least 6 digits
 *      6   354294  6   // this is our upper bound
 */
function sumOfFifthPowerSums() {
    "use strict";
    var n, s, i, temp, sum = 0;
    
    for (n = 10; n <= 354294; n += 1) {
        s = String(n);
        temp = 0;
        for (i = 0; i < s.length; i += 1) {
            temp += Math.pow(parseInt(s[i], 10), 5);
        }
        if (temp === n) {
            console.log(n);
            sum += n;
        }
    }
    return sum;
}

/* Problem 31
 * In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:
 *      1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
 * It is possible to make £2 in the following way:
 *      1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
 * How many different ways can £2 be made using any number of coins? */

/* I want something along the lines of
 * if 200 -> 100, 100
 * if 100 -> 50, 50
 * if  50 -> 20. 20, 10
 * if  20 -> 10, 10
 * if  10 -> 5, 5
 * if   5 -> 2, 2, 1
 * if   2 -> 1, 1
 * if   1 -> end
*/
function coinPossibilities(end) {
    "use strict";

    var ways = 0, a, b, c, d, e, f, g;
    
    // sanatize input
    end = parseInt(end, 10);
    if (isNaN(end)) {
        end = 200;
    }
 
    for (a = end; a >= 0; a -= 200) {
        for (b = a; b >= 0; b -= 100) {
            for (c = b; c >= 0; c -= 50) {
                for (d = c; d >= 0; d -= 20) {
                    for (e = d; e >= 0; e -= 10) {
                        for (f = e; f >= 0; f -= 5) {
                            for (g = f; g >= 0; g -= 2) {
                                ways += 1;
                                if (ways % 10000 === 0) {
                                    console.log(ways);
                                }
                            }
                        }
                    }
                }
            }
        }
    }
    return ways;
}

// sorted array of ints, return array w/o duplicates
11123345