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.

Understanding the Problem

For an N-digit number, we need to:

• Calculate sum of digits at even positions (0-indexed)
• Calculate sum of digits at odd positions (0-indexed)
• Check if even position sum is divisible by A
• Check if odd position sum is divisible by B

Helper Function: Calculate Position-wise Sums

First, let's create a function to calculate sums based on digit positions:

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};
}; 

// Test the helper function
console.log("Testing indexSum for number 1234:");
console.log(indexSum(1234));

Testing indexSum for number 1234:
{ sumOdd: 6, sumEven: 4 }

Divisibility Check Function

const divides = (b, a) => a % b === 0;

// Test divisibility
console.log("Testing divisibility:");
console.log("15 divisible by 5:", divides(5, 15));
console.log("17 divisible by 5:", divides(5, 17));

Testing divisibility:
15 divisible by 5: true
17 divisible by 5: false

Main Function: Count Valid Numbers

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;
};

// Example: Find 2-digit numbers where even position sum divisible by 5 and odd position sum divisible by 3
console.log("2-digit numbers (A=5, B=3):");
console.log(countNum(2, 5, 3));

2-digit numbers (A=5, B=3):
[ 30, 35, 60, 65, 90, 95 ]

Complete Example with Explanation

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;
};

// Test with different parameters
console.log("N=2, A=5, B=3:", countNum(2, 5, 3).length, "numbers found");
console.log("N=2, A=2, B=3:", countNum(2, 2, 3).length, "numbers found");

N=2, A=5, B=3: 6 numbers found
N=2, A=2, B=3: 30 numbers found

How It Works

For number 35:

• Digits: 3 (position 1, odd), 5 (position 0, even)
• Even position sum: 5
• Odd position sum: 3
• 5 divisible by 5: ?, 3 divisible by 3: ?

Conclusion

This solution recursively calculates position-wise digit sums and filters N-digit numbers based on divisibility conditions. The algorithm has O(10^N) time complexity as it checks all N-digit numbers.