Finding all the n digit numbers that have 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. Let's say the three numbers are a, b and n.

Our job is to find all the n-digit numbers whose sum of digits at even positions and odd positions are divisible by a and b respectively. And we have to lastly return an array containing all the required numbers, the array should be empty if there are no matching numbers.

Example

Following is the code −

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(3, 5, 3));

Output

This will produce the following output in console −

[    104, 109, 134, 139, 164, 169, 194, 199, 203,    208, 233, 238, 263, 268, 293, 298, 302, 307,    332, 337, 362, 367, 392, 397, 401, 406, 431,    436, 461, 466, 491, 496, 500, 505, 530, 535,    560, 565, 590, 595, 604, 609, 634, 639, 664,    669, 694, 699, 703, 708, 733, 738, 763, 768,    793, 798, 802, 807, 832, 837, 862, 867, 892,    897, 901, 906, 931, 936, 961, 966, 991, 996 ]