8) Largest product in a series

Find the greatest product of five consecutive digits in the 1000-digit number.

by fosco

JavaScript

var numberSequence = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450',
    offset = 5,
    strLength = numberSequence.length,
    n = parseInt(strLength / offset, 10),
    bigger = {
        'value': 0,
        'mult': 0
    };

for (var i = 0; i < offset; i++) {
    for (var j = 0; j < n; j++) {
        var position = j * offset + i;
        if (position + offset > strLength) {
            break;
        }

        var value = numberSequence.substr(position, offset),
            mult = 1;

        for (var k = 0; k < offset; k++) {
            mult *= parseInt(value[k], 10);
        }

        if (mult > bigger.mult) {
            bigger = {
                'value': value,
                'mult': mult
            };
        }
    }
}

console.log(bigger); //{value: "99879", mult: 40824}