1. Total number of seven-digit numbers

A seven-digit number ranges from $1000000$ to $9999999$. So the first digit can be chosen in $9$ ways ($1$ to $9$), and each of the remaining $6$ digits can be chosen in $10$ ways.

Hence total seven-digit numbers: $9\cdot 10^6$

2. We want numbers whose digit sum is even

Let the number be $d_1d_2d_3d_4d_5d_6d_7$ with $d_1\in\{1,2,\dots,9\}$ and $d_2,\dots,d_7\in\{0,1,\dots,9\}$.

We count how many such numbers have even digit sum.

3. Count even/odd choices for each digit

• For the first digit $d_1$ (from $1$ to $9$):

  • odd choices: $1,3,5,7,9$ $\Rightarrow 5$
  • even choices: $2,4,6,8$ $\Rightarrow 4$

• For each of the last $6$ digits (from $0$ to $9$):

  • odd choices: $5$
  • even choices: $5$

4. Count by parity of the last 6 digits

Since each of the last $6$ digits has equal numbers of even and odd choices, among all $10^6$ choices for $(d_2,\dots,d_7)$, exactly half have even sum and half have odd sum.

So:

• number of 6-digit tails with even sum $= \dfrac{10^6}{2}=5\cdot 10^5$
• number of 6-digit tails with odd sum $= 5\cdot 10^5$

5. Form total numbers with even digit sum

For the total sum to be even:

• first digit even and last 6 digits sum even, or
• first digit odd and last 6 digits sum odd.

Therefore required count is $4\cdot 5\cdot 10^5 + 5\cdot 5\cdot 10^5$ $=(4+5)\cdot 5\cdot 10^5$ $=9\cdot 5\cdot 10^5$ $=45\cdot 10^5$

This matches the form $m\cdot n\cdot 10^n$ with $m=9,\quad n=5$

Thus, $m+n=9+5=14$

6. Comparison with stored answer

Derived answer: $14$

Stored correct answer: $14$

They agree.