Count of N-digit Numbers having Sum of even and odd positioned digits divisible by given numbers - JavaScript

We are required to write a JavaScript function that takes in three numbers A, B and N, and finds the total number of N-digit numbers whose sum of digits at even positions and odd positions are divisible by A and B respectively.

Example

Let’s write the code for this function −

const indexSum = (num, sumOdd = 0, sumEven = 0, index = 0) => {
   if(num){
       if(index % 2 === 0){
           sumEven += num % 10;
       }else{
           sumOdd += num % 10;
       };
 
       return indexSum(Math.floor(num / 10), sumOdd, sumEven, ++index);
   };
   return {sumOdd, sumEven};
 
}; 
const divides = (b, a) => a % b === 0;
const countNum = (n, first, second) => {
   let start = Math.pow(10, (n-1));
   const end = Math.pow(10, n)-1;
   const res = [];
   while(start <= end){
       const { sumEven, sumOdd } = indexSum(start);
       const condition = divides(first, sumEven) && divides(second, sumOdd);
       if(condition){
           res.push(start);
       };
       start++;
   };
   return res;
};
console.log(countNum(2, 5, 3));

Output

Following is the output in the console −

[ 30, 35, 60, 65, 90, 95 ]