Sum up a number until it becomes 1 digit JS

Question

Before you mark duplicate note that it isn't. Others didn't look for the exact same thing as me.

What is the most compact possible way to sum up a number in javascript until there is only one digit left. For example: You input 5678 then the script adds it together (5+6+7+8) and gets 26, but since its more than 1 digit it adds it again and gets 2+6=8.

Is there anyway to do this with a number of any size? And how compact can the script get?

Is this meant to be [code-golfing](https://codegolf.stackexchange.com/) ? — Sirko, Apr 18 '18 at 06:19
Possible duplicate of [Sum all the digits of a number Javascript](https://stackoverflow.com/questions/38334652/sum-all-the-digits-of-a-number-javascript) — , Apr 18 '18 at 06:20
no it isn't. The answers in that question are different and only get down to 2 digits. — VinceKaj, Apr 18 '18 at 06:21
@Vincent [This answer is only down to 2 digits?](https://stackoverflow.com/a/47486307/47589) — , Apr 18 '18 at 06:22

Mark · Accepted Answer · 2018-04-18T07:06:02.963

27

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.

edited Apr 18 '18 at 07:06

answered Apr 18 '18 at 06:35

Mark

90,562
7
108
148

2

cool! This is so smart way. While my approach looks more functional programming style :) – Dong Nguyen Apr 18 '18 at 06:39
@Titus any idea how to fix it? – Dong Nguyen Apr 18 '18 at 06:40
It seems to fail when the number is divisible by `9`. Probably to fix that, it will get more complicated then other solution. – Titus Apr 18 '18 at 06:42
@Titus, is there a case that isn't fixed with the simple edit? – Mark Apr 18 '18 at 06:44
I can't think of any and it seems to be logical since the result can never be `0` – Titus Apr 18 '18 at 06:46
@vincent, can you give an example? A few cases I've tried: 920 yields 2 which is correct, 931 yields 4, also correct. – Mark Apr 18 '18 at 06:51
9999 or 1234567890 which is the number I need to test – VinceKaj Apr 18 '18 at 06:55
@Vincent `9+9+9+9` == `36` `3+6` == `9` that's the answer I get. Same with `1+2+3+4+5+6+7+8+9+0` = `45` `4+5` ==`9` Also what I get. Are you sure you're getting the `n || 9` in the `console.log()`? – Mark Apr 18 '18 at 06:59
9999 % 9 = 0. That's the flaw. Even if it gets to 9+9+9+9=36, 3+6=9, 9 % 9 = 0. It console logs 0. – VinceKaj Apr 18 '18 at 07:01
@Vincent double check the last line — that's addressed. – Mark Apr 18 '18 at 07:02
@Mark_M do you think that `while` loop is really required ?? – ashwintastic Apr 18 '18 at 07:04
@ashwintastic no - of course not — that's really funny that I didn't notice that. Thanks. – Mark Apr 18 '18 at 07:05

score 8 · Answer 2 · answered Apr 18 '18 at 06:21

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)));

score 2 · Answer 3 · edited Nov 19 '19 at 10:37

2

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

edited Nov 19 '19 at 10:37

fonZ

2,428
4
21
40

answered Oct 21 '19 at 13:45

Foysal Ahmed Emon

67
6

2

Please describe, what was the problem and how will this code solve it, to help others understand this answer – FZs Oct 21 '19 at 17:13

score 2 · Answer 4 · edited Nov 28 '19 at 19:31

2

function createCheckDigit(membershipId) {
     var total = membershipId;
   while (total.length > 1) {
       t = 0;
       for (let i = 0; i < total.length; i++) {
               t = Number(t) + Number(total[i]);
       }
       total = t.toString(); 
   }
   return total;
  }
console.log(createCheckDigit("55555"));

edited Nov 28 '19 at 19:31

Dhruv Narayan Singh

647
9
21

answered Nov 28 '19 at 18:04

Vinod Saggam

21
2

When answering an old question, your answer would be much more useful to other StackOverflow users if you included some context to explain how your answer helps, particularly for a question that already has an accepted answer. See: [How do I write a good answer](https://stackoverflow.com/help/how-to-answer). – David Buck Nov 28 '19 at 18:24

Nina Scholz · Answer 5 · 2018-04-18T06:48:32.053

1

For a compact way, you could use an iterative and recursive approach.

const d = n => n > 9 ? d([...n.toString()].reduce((a, b) => +a + +b)) : n;

console.log(d(5678));

edited Apr 18 '18 at 06:48

answered Apr 18 '18 at 06:41

Nina Scholz

376,160
25
347
392

gurvinder372 · Answer 6 · 2018-04-18T06:26:57.980

You need to check if the length of value (after converting sum to a String) is 1 or not.

var fnSum = (arr) => arr.reduce((a, c) => a + (+c), 0); //function to get sum of values

var fnGetSingleDigit = (arr) => {
  while (arr.length != 1) {
    var sum = fnSum(arr);
    arr = String(sum).split("");
  }
  return arr[0];
};

console.log(fnGetSingleDigit([5, 6, 7, 8]));

console.log(fnGetSingleDigit([41,34,3,54,63,46]));

Demo

var fnSum = (arr) => arr.reduce((a, c) => a + (+c), 0); //function to get sum of values

var fnGetSingleDigit = (arr) => {
  while (arr.length != 1) {
    var sum = fnSum(arr);
    arr = String(sum).split("");
  }
  return arr[0];
};

console.log(fnGetSingleDigit([5, 6, 7, 8]));

console.log(fnGetSingleDigit([41,34,3,54,63,46]));

score 0 · Answer 7 · answered Apr 18 '18 at 06:32

function fun(n) {
sum = 0;
while(n > 0 || sum > 9)
    {
        if(n == 0)
        {
            n = sum;
            sum = 0;
        }
        sum = sum+n % 10;
        n = Math.floor(n/10);
    }
    console.log(sum);
}

You can use the above snippet.

Parikshit Sarkar · Answer 8 · 2020-02-17T11:53:58.307

0

I did it like this:

function sumOfID(__id){
  var arr_str = Array.from(__id)
  var arr = arr_str.map((t)=>{return parseInt(t)})
  function digSum(__arr){
   return __arr.reduce((a,b)=>{return a + b})
  }
   return digSum(arr)%9 || 9    
 }

Note : The trick here is to use mod9 because the modulo 9 of any multiple digit number is its total sum until it becomes a single digit number.

edited Feb 17 '20 at 11:53

answered Feb 17 '20 at 11:35

Parikshit Sarkar

570
5
6

emre-ozgun · Answer 9 · 2021-07-07T23:48:36.083

// recursive
    function digitSum(n) {
      return (n % 10 === n) ? n : (digitSum(n.toString().split('').reduce((acc, curr) => acc + parseInt(curr), 0)));
    }

    
        
   function digitSum(n) {
       while(!(n % 10 === n)) {
         n = n.toString().split('').reduce((acc, curr) => acc + parseInt(curr), 0);
       }
      
       return n;
    
    console.log(digitSum(467)); // -> 17 -> 8

basically check if the nums length (digit count length) < 2 (achieved this with n % 10 === n (9 % 10 === 9) or n.toString().length < 2 if NOT then call the function again with calculated digit sum take example 467:

4 + 6 + 7 = 17 (digit length is not 1) call with 17
1 + 7 = 8 (digit length is 1) -> return. 2nd version is the same as recursive one, hope this helps!

While this code may answer the question, providing additional context regarding *how* and/or *why* it solves the problem would improve the answer's long-term value. — Sven Eberth, Jul 07 '21 at 22:24

score 0 · Answer 10 · answered Oct 17 '21 at 10:01

0

const a = [1,2,4,4,5,6,7,7].join('')

function add(i){
  let x = i.toString().split('').reduce((p,c)=>parseInt(p)+parseInt(c))
  x = x.toString().length > 1 ? add(x) : x
  return parseInt(x)
}


console.log(add(a))

answered Oct 17 '21 at 10:01

ramVermilion

51
3

score -1 · Answer 11 · answered Apr 18 '18 at 06:20

-1

Here you go

var number = 56785;
var total = number + '';
    
while(total.length > 1) {
   var temp = 0;
   for(var i = 0; i < total.length; i++) {
       temp += +total[i];
   }
   total = temp + '';
}
    
console.log(total);

answered Apr 18 '18 at 06:20

Bunyamin Coskuner

8,719
1
28
48

score -1 · Answer 12 · answered Apr 18 '18 at 06:23

You can also opt to use the forEach loop and a while condition to loop over the stringify number and get the sum until the length is 1.

var num = 5678;
var strNum = num.toString().split('');
while(strNum.length !== 1){
  var strNum = calculateSum(strNum);
  strNum = strNum.split('');
}
var result = parseInt(strNum);
console.log(result);

function calculateSum(strNum){
  var sum = 0;
  strNum.forEach((digit)=>{
    sum += parseInt(digit);
  });
  return sum.toString();
}

score -1 · Answer 13 · answered Apr 18 '18 at 06:26

-1

This is my solution for the same problem on CodeWars

const sumup = (num) => {
   let a = String(num).split('').map((k) => {
     return Number(k);
   }).reduce((p, c) => {
     return p + c;
   }, 0);

   return a > 9 ? sumup(a) : a;
}

answered Apr 18 '18 at 06:26

Dong Nguyen

1,239
9
17

score -1 · Answer 14 · answered Apr 18 '18 at 07:07

Here is a recursive solution without using any library!

function getSumOfDigits(num){
  var result = 0;
  while(~~(num/10) > 0){
    result += (num % 10);
    num = ~~(num/10);
  }
  return result + num;
}

function getSingleDigit(num) {
  if(~~(num/10) === 0) {
    return num
  } else {
    num = getSumOfDigits(num);
    return getSingleDigit(num);
  }
}

console.log(getSingleDigit(393));

score -1 · Answer 15 · answered Aug 02 '18 at 00:43

You can combine some utility methods with recursion and do it in two lines that are still pretty readable:

function sumDigits(digits) {
  let sum = Array.from(digits.toString()).map(Number).reduce((res, val) => res + val, 0);
  return sum < 10 ? sum : sumDigits(sum);
}

console.log(sumDigits(55555)) // 5+5+5+5+5 = 25 -> 2+5 = 7

score -1 · Answer 16 · answered Jul 18 '19 at 08:08

-1

function doSum(n){
 var temp = n, total= 0;
  for(var i = 0; i < n.toString().length; i++ ){
        total = total + (Math.floor(temp) % 10)
    temp = (temp / 10)
  }
 if(total > 9 ) { return doSum(total) }
return total
}
doSum(999999)

answered Jul 18 '19 at 08:08

Mohammed Jafar

493
6
4

Sum up a number until it becomes 1 digit JS

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