The number of 5-digit numbers such that their sum of digits is even is? I tried this problem by counting the number of 5-digit numbers and then dividing it by 2 just like we do for three digits. Please help. Thank you.

Can you see how many $5$ digit nos. can be formed with nos. $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$? 
Clearly $1$st place can't be occupied by $0$, so the $1$st place can be filled from $9$ different numbers.   The rest of the places can be filled from $10$ different numbers, so the total number of five digit numbers that can be formed is $9\times10\times10\times10\times10$. 
Now these many digits will be either even or odd after summation.  But, we are interested in the numbers whose sum is even so it will be half of the total five digit numbers, so the answer is $\frac{1}{2}\times(9\times10\times10\times10\times10)$ which is $45000$.
