Question

How many different 7 digit numbers are there the sum of whose digits is even? The answer is expressed as $k\times {{10}^{5}}$. Find the value of k.

Answer 1

Hint: Consider numbers of the form ${{a}_{1}}{{a}_{2}}{{a}_{3}}{{a}_{4}}{{a}_{5}}{{a}_{6}}i$ where $i$ can take the values $0-9$. Count the number of possible values for each of ${{a}_{i}}$. Count the possible values of $i$ for a given set of numbers ${{a}_{1}}-{{a}_{6}}$. Multiply all the values to get the total number of seven digit numbers which have even sum.

Complete step-by-step answer:
We have to count the possible number of 7 digit numbers which can be formed such that the sum of digits is even.
We will consider numbers of the form ${{a}_{1}}{{a}_{2}}{{a}_{3}}{{a}_{4}}{{a}_{5}}{{a}_{6}}i$, where $i$ can have any of the digits from $0-9$.
So, the possible seven digit numbers are of the form ${{a}_{1}}{{a}_{2}}{{a}_{3}}{{a}_{4}}{{a}_{5}}{{a}_{6}}0,{{a}_{1}}{{a}_{2}}{{a}_{3}}{{a}_{4}}{{a}_{5}}{{a}_{6}}1,...{{a}_{1}}{{a}_{2}}{{a}_{3}}{{a}_{4}}{{a}_{5}}{{a}_{6}}9$.
We observe that ${{a}_{1}}$ can have any of the digits from $1-9$, while ${{a}_{2}}-{{a}_{6}}$ can have any of the digits from $0-9$.
So, the number of digits possible at ${{a}_{1}}$ are 9 and number of digits possible at each of ${{a}_{2}}-{{a}_{6}}$ is 10.
For a fixed value of ${{a}_{1}}-{{a}_{6}}$, $i$ can take only odd values or even values. So the number of possible values of $i$ for fixed values of ${{a}_{1}}-{{a}_{6}}$ is 5.
To calculate the number of possible seven digit numbers which have an even sum of digits, multiply the number of possible digits which can be placed at each place.
First place from right side i.e ${{a}_{1}}$ place, numbers can be filled in 9 ways and ${{a}_{2}}-{{a}_{6}}$ numbers can be filled in $10\times10\times10\times10\times10$ ways (as repetition is allowed) and last digit can be filled in 5 ways.
Thus, the number of possible seven digit numbers which have an even sum of digits $=9\times10\times10\times10\times10\times10\times 5$
$=9\times {{10}^{5}}\times 5$.
So, the number of possible seven digit numbers which have even sum of digits $=45\times {{10}^{5}}$.
We know that the number of possible seven digit numbers which have an even sum of digits expressed as $k\times {{10}^{5}}$.
Thus, we have $k\times {{10}^{5}}=45\times {{10}^{5}}$.
So, we have $k=45$.
Hence, the value of k is 45.

Note: We can also solve this question by calculating the total number of seven digits numbers which can be formed and dividing it by 2 because half of them will have odd sum and half of them will have even sum. This is a shorter trick to solve the question and it will also save our time.