JS Digital Root

digitalroot.js

function root(num){
    var total = 0;
    if(num.toString().length == 1){
        var iNum = parseInt(num);
        return iNum;
    }else{
        num.toString().split("").forEach( function(value){
            var iValue = parseInt(value);
            return total += iValue;
        });
        return root(total);
    }
}

this was the best answer
function digital_root(n) {
return (n - 1) % 9 + 1;
}

n-1 means the last element
% modulus for the remainder

function digitalRoot(num) {
if (num < 10 ) return num;
return digitalRoot(Math.floor(num/10)) + num % 10;
}

came across something in a Python coding challenge. If n == 0 then that equation will return 9

def digital_root(n):
    if n == 0:
        return 0
    else:
        return (n-1) % 9 + 1

print(digital_root(0), 0)
print(digital_root(16), 7)

also in Ruby...

def digital_root(n)
  if n === 0
    return 0
  else
    return (n-1) % 9 + 1
  end
end

function digitalRoot(n) {
return n % 9;
}
that's it