# About this series

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.

## Advanced developers

Please provide alternative solutions in the comments below.

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)
{
sum = add(sum, r);
}
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;
}
function add(sNumber1, sNumber2)
{
var maxSize = Math.max(sNumber1.length, sNumber2.length);
var s1 = sNumber1.padStart(maxSize, "0");
var s2 = sNumber2.padStart(maxSize, "0");
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)