TASK MUST BE ADDED TO THE GIVEN PROGRAM BELOW. PLEASE DON'T WRITE THE CODE AS A SEPERATE PROGRAM!

PLEASE READ THE TASK CAREFULLY BEFORE YOU START. IF YOU ARE NOT SURE WHAT TO DO, PLEASE SKIP MY QUESTION. I WILL GIVE NEGATIVE RATING FOR WRONG OR INCOMPLETE ANSWER.

THANK YOU.

TASK :

A well-known theorem of math states that the above sum is 9 iff the number is divisible by 9. Write a function that modifies its argument number so that its divisible by 9. The function should add something to the right-most digit of the number if possible; otherwise it should subtract something from that digit. An example use of this function might be:

transformNum(n); cout << n; // prints 234567 if n was originally 234565

Write the transformNum function. You should be able to reuse most of the code from earlier. Also modify the driver program to output the transformed number.

// GIVEN PROGRAM STARTS HERE

#include

using namespace std;

// Precondition: num > 0

// Postcondition: the return number of digits in num

int numDigits(int num);

// Precondition: num > 0, index > 0

// Postcondition: return the index'th digit of number

int getDigit(int num, int index);

// Precondition: num > 0

// Postcondition: the return value is the iterated sum of digits of number

int sumDigits(int num);

int main(){

int n;

cout<<"Enter a number: ";

cin>>n;

cout<<"The Sum of digits: "<

return 0;

}

int numDigits(int num){

int count = 0;

while(num > 0){

num = num/10;

count++;

}

return count;

}

// Precondition: num > 0, index > 0

// Postcondition: return the index'th digit of num

int getDigit(int num, int index){

int digit;

for(int i=1; i<=index; i++){

digit = num%10;

num = num/10;

}

return digit;

}

// Precondition: num > 0

// Postcondition: the return value is the iterated sum of digits of num

int sumDigits(int num){

int n = numDigits(num);

int sum = 0;

for(int i=1; i<=n; i++)

sum = sum + getDigit(num, i);

return sum;

}