Question

1. Write a C function called weighted_digit_sum that takes a single integer as input, and returns a weighted sum of that numbers digits. The last digit of the number (the ones digit) has a weight of 1, so should be added to the sum "as is". The second from last digit (the tens digit) has a weight of 2, and so should be multiplied by 2 then added to the sum. The third from last digit should be multiplied by 1 and added to the sum, and so on, with the weights alternating between 1 and 2 for each digit in turn. The final sum of all weighted digits should be returned from the function. For example, if the number 2561 was input to the function, the sum would be 1x1 + 2x6 + 1x5 + 2x2 = 22. HINT: Start from the right-most digit, and use simple C operators to find both the digit and the remaining number once that digit is removed.
2. Write another C function called reduced_sum which repeatedly calls the weighted_sum function developed in part (a) until the result is a single digit (less than 10). This value is then returned from the function. For example, if the same number 2561 was input to the function, it would return 4, since 2561 -> 22 -> 6 (1x2 + 2x2). Therefore 6 will be returned.
3. The only possible results of the reduced_sum function are 0 to 9. Write a main() program which uses the functions developed in parts (a) and (b) to find how many times each of these possible results occurs for the numbers from 0 to 10000. You should not print out the result of each test, just the total sum for each of the final results 0 through 9, so 10 total lines of output.

Answer 1

Code:

#include <stdio.h> // for accessing the standard input/output

// implementing the weighted digit function
int weighted_digit_sum(int n){
    int digit=0;
    int sum=0;
    int count=1;
    while(n){   //we use a while loop to get and apply added weight to it
        digit = n%10;   // getting the rightmost digit
        if(count%2==0){     // if it falls on 2nd place, multiply by 2 or else 1
            sum += digit*2;
        }else{
            sum += digit*1;
        }
        n=n/10;
        count++;
    }
    return sum; // returning the weighted value
}

// implementing the reduced sum function
int reduced_sum(int x){
    int y=x;
    while(y>9){  // condition break the loop when sum<10 as we need only single digit sum
        y=weighted_digit_sum(y);
    }
    return y;
}

// driver code
int main() {
    int sm = 0;
        int a[10];
        for(int i=0; i<=10000; i++){
            sm=reduced_sum(i);
            a[sm]++;
        }
        for(int j=0; j<10; j++){
            printf("%d\n", a[j]);
        }
        
        return 0;
}

Output: