# Recursive product of summed digits JavaScript

We have to create a function that takes in any number of arguments (Number literals), adds them together, and returns the product of digits when the answer is only 1 digit long.

For example −

If the arguments are −

16, 34, 42

We have to first add them together −

16+34+42 = 92

And then keep multiplying the digits together until we get a 1-digit number like this −

9*2 = 18
1*8 = 8

When we get the one-digit number, we have to return it from our function.

We will break this into two functions −

• One function accepts a number and returns the product of its digits, we will use recursion to do so, let’s call this first function product().

• Second function recursively calls this product() function and checks if the product happens to be 1 digit, it returns the product otherwise it keeps on iterating.

The code for this whole functionality will be −

## Example

const recursiveMuliSum = (...numbers) => {
const add = (a) => a.length === 1 ? a[0] : a.reduce((acc, val) => acc+val);
const produce = (n, p = 1) => {
if(n){
return produce(Math.floor(n/10), p*(n%10));
};
return p;
};
if(res > 9){
return recursiveMuliSum(res);
}
return res;
};
console.log(recursiveMuliSum(16, 28));
console.log(recursiveMuliSum(16, 28, 44, 76, 11));
console.log(recursiveMuliSum(1, 2, 4, 6, 8));

## Output

The output in the console will be −

6
5
2

Updated on: 24-Aug-2020

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