Question

Find the number of seven-digit numbers which can be formed with the sum of the digits being even.

Answer 1

Find the number of seven-digit numbers.

we know that even numbers have $0,2,4,6,8$ at unit position.

Let,

$9$

$10$

$10$

$10$

$10$

$10$

$5$

To make seven-digit numbers,
In the first place of the box, we can have $9$ digits to fill from $1$ to $9$ . We have keep out $0$, because $0$ cannot be placed in the first position of the box. If we put $0$, then it will become a $6$ digit number.

That's why in the first place, we have $9$ ways to fill the box.

We can fill $10$ digits from $0$ to $9$ in the second place.

In the third position of the box, we can fill in $10$ digits from $0$ to $9$, and in the fourth place, we can fill $10$ digits from $0$ to $9$.

Similarly, we can fill $10$ digits from $0$ to $9$ in the fifth and sixth places, and we have $5$ digits to fill at the end of the unit place, which are $0,2,4,6,8$.

Consequently, the total number of seven – digit numbers the sum of whose digits is even is,

Required numbers $=9\times 10\times 10\times 10\times 10\times 10\times 5$

$=45\times {10}^5$

Hence, the number of seven-digit numbers which can be formed with the sum of the digits being even is $45\times {10}^5$.