How many different seven-digit numbers are there such that the sum of whose digits is even?

Hint: We can choose any number from (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9), to fill the seven places of the required 7 digit number. The first digit can’t be zero so find all possible 7 digit numbers. Use these total possible 7-digit numbers formed to find those whose sum of digits is even.

Complete step-by-step answer:

Total number of possible digits are (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9).
So the total digits are 10.
Now we have to make a seven digit number.
So the first place is filled by 9 ways as the number cannot start by 0, otherwise the seven digit number becomes a six digit number.
Now as repetition is allowed so the other digits are filled by 10 ways.
So the total number of seven digit number = $9 \times {\left( {10} \right)^6}$
Now as we know that after even numbers there is an odd number or vice-versa.
And the sum of the digits are either even or odd.
For example the first seven digit number is 1000000, so the sum of digits is 1, which is odd.
And the next seven digit number is 1000001, so the sum of digits is 2, which is even.
So we can say that from the total numbers the seven digit numbers whose digit sum is even is half of the total numbers.
Therefore, the different seven digit numbers whose digit sum is even $ \Rightarrow \dfrac{{9000000}}{2} = 4500000 = 45 \times {\left( {10} \right)^5}$
So this is the required answer.

Note: There is a misconception that if the number is an even number then its sum of digits should be even only, or if the number is an odd number then the sum of digits of it will be odd only. For example 18 is an even number however its sum of digits is 9 which is odd.