Q: I need to write a program that asks the user to input a positive integer n, and print all of the numbers from 1 to n that have more even digits than odd digits.

For example, if n=30, the program should print:

2

4

6

8

20

22

24

26

28

currently my code looks like this:

#include  using namespace std; int main() { int n, i; int even_count = 0; int odd_count = 0; cout << "Please enter a positive integer: "; cin >> n; for (i = 1; i < n; i++) { if(i % 2 == 0) even_count++; else odd_count++; } cout << even_count << endl; cout << odd_count << endl; } 

I know that I need to check each number between 1 and n to see if the number has more even than odd numbers. if it does have more even numbers, then I need to print it to the console, if not then skip. Doing this all the way up to n. However, I just can't seem to figure it out.

As my code is right now. it will loop through every number between 1 and n, which I want to do, but instead of telling me how many odds and evens I have for each number between 1 and n it just tells me how many I have for n.

How do I check each number so that I can compare odds against evens? I thought If I add in a nested if statement like:

if (even_count > odd_count) { cout << i << endl; } 

that it might do the trick, but I was mistaken.

I know there is a gap in my code I just cant get it! Any help is appreciated.

Thank you!