If the number of seven-digit numbers, such that the sum of their digits is even, is $m \cdot n \cdot 10^k$; $m, n \in \{1, 2, 3, \dots, 9\}$, then $m + n$ is equal to
Solution:
A seven-digit number can be represented as $d_1 d_2 d_3 d_4 d_5 d_6 d_7$.
For a number to be a valid seven-digit number, the leading digit $d_1$ cannot be zero.
So, $d_1 \in \{1, 2, 3, \dots, 9\} \Rightarrow 9$ choices.
The next five digits $d_2, d_3, d_4, d_5, d_6$ can be any digit from 0 to 9.
So, each of these 5 digits has 10 choices.
The sum of all seven digits is $S = d_1 + d_2 + d_3 + d_4 + d_5 + d_6 + d_7$.
We are given that the sum $S$ must be even. Let’s analyze the last digit based on the sum of the first six digits.
Case 1: If the sum of the first six digits $(d_1 + d_2 + \dots + d_6)$ is even, then the last digit $d_7$ must be even for the total sum $S$ to be even.
Possible even digits for $d_7$ are $\{0, 2, 4, 6, 8\} \Rightarrow 5$ choices.
Case 2: If the sum of the first six digits is odd, then the last digit $d_7$ must be odd for the total sum $S$ to be even.
Possible odd digits for $d_7$ are $\{1, 3, 5, 7, 9\} \Rightarrow 5$ choices.
In either case, regardless of what the first six digits sum up to, there are exactly 5 valid choices for the last digit $d_7$.
Therefore, the total number of such seven-digit numbers is:
$$9 \times 10 \times 10 \times 10 \times 10 \times 10 \times 5$$
$$= 9 \times 10^5 \times 5$$
$$= 45 \times 10^5$$
$$= 9 \cdot 5 \cdot 10^5$$
Comparing this with the given form $m \cdot n \cdot 10^k$, where $m, n \in \{1, 2, \dots, 9\}$:
$$m = 9, \quad n = 5, \quad k = 5$$
The question asks for the value of $m + n$:
$$m + n = 9 + 5 = 14$$
Ans. (14)