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# What’s your alternative solution? Challenge #59

This is series of daily JavaScript coding challenges... for both beginners and advanced users.

Each day I’m gone present you a very simple coding challenge, together with the solution. The solution is intentionally written in a didactic way using classic JavaScript syntax in order to be accessible to coders of all levels.

Solutions are designed with increase level of complexity.

## Today’s coding challenge

``````Calculate 70! with high precision (all digits)

``````

(scroll down for solution)

## Code newbies

If you are a code newbie, try to work on the solution on your own. After you finish it, or if you need help, please consult the provided solution.

You can solve it using functional concepts or solve it using a different algorithm... or just solve it using the latest ES innovations.

By providing a new solution you can show code newbies different ways to solve the same problem.

## Solution

``````// Solution for challenge53

println(factorial(70));

// Calculate factorial(n) ... using big number calculations
function factorial(n)
{
var prod = "1";

for(var i = 2; i <= n; i++)
{
prod = mult(prod, i.toString());
}

return prod;
}

// Multiplies sNumber1 * sNumber2
// Each number is provided as string
function mult(sNumber1, sNumber2)
{
// Calculate partial results according to multiplication algorithm
var partialResults = [];

for(var i = sNumber2.length - 1; i >= 0; i--)
{
var digit = parseInt(sNumber2[i]);

var partialResult = multDigit(sNumber1, digit);
partialResult += "0".repeat(partialResults.length);

partialResults.push(partialResult);
}

// Sum partial results to obtain the product
var sum = "";

for(var r of partialResults)
{
}

return sum;
}

// Multiplies number sNumber (as string) with a single digit number
function multDigit(sNumber, digit)
{
var p = "";
var carry = 0;

for(var i = sNumber.length - 1; i >= 0; i--)
{
var numberDigit = parseInt(sNumber[i]);

var prod = digit * numberDigit + carry;
var prodDigit = prod % 10;
carry = Math.floor(prod / 10);

p = prodDigit.toString() + p;
}

if (carry > 0)
p = carry + p;

return p;
}

{
var maxSize = Math.max(sNumber1.length, sNumber2.length);

var s = "";
var carry = 0;

for(var i = maxSize - 1; i >= 0; i--)
{
var digit1 = parseInt(s1[i]);
var digit2 = parseInt(s2[i]);

var sum = digit1 + digit2 + carry;
var digitSum = sum % 10;
carry = sum >= 10 ? 1 : 0;

s = digitSum.toString() + s;
}

if (carry > 0)
s = carry + s;

return s;
}

``````

To quickly verify this solution, copy the code above in this coding editor and press "Run".

Note: The solution was originally designed for codeguppy.com environment, and therefore is making use of `println`. This is the almost equivalent of `console.log` in other environments. Please feel free to use your preferred coding playground / environment when implementing your solution.

## Discussion (0) 