Sum up a number until it becomes 1 digit JS

If you're looking for short, it's hard to beat:

var n = 5678;
sum  = n % 9 || 9;

console.log(sum)

If you're curious about how that works, see: casting out nines.

You can do this using recursion, here is an example:

function getOneDigit(nr){
    let asStr = nr.toString();
    let sum = asStr.split("").reduce((a, c) => {
         a+=parseInt(c);
         return a;
    }, 0);
    return sum >= 10 ? getOneDigit(sum) : sum;
}

[234235, 235346, 657234, 1, 2, 5423].forEach(nr => console.log(getOneDigit(nr)));

This line of code does this.

function digital_root(n) {
    return (n - 1) % 9 + 1;
}

When you take the modulo of a number with 9, it becomes a 1 less than 9. So, when you have 10%9, you will get 1, which is equal to 1+0.

Similarly, 123%9 = 6.

Why (n-1)%9 and added 1: Because when you input 9, it became 0 so input will be invalid.

For n=9, this method will return (9-1)%9 = 8 and 8+1 = 9.

So it does not break.

You can also write

return n%9 || 9