If the number of seven-digit numbers, such that the sum of their digits is even, is m * n * 10^n; m,

Q: Wait, why can't we just say half of all 7-digit numbers have an even sum? Is it always exactly 50%?

That is a brilliant intuition! For large sets of numbers, the parity is indeed balanced. However, in combinatorics, we must prove it. By showing that for every possible sum of the first six digits, there are exactly 5 choices for the last digit, we mathematically guarantee that exactly half of the total numbers satisfy the condition.

Q: Does the restriction on the first digit ($d_1 \neq 0$) mess up the parity argument?

Not at all! The parity argument relies on the last digit $d_7$ 'fixing' the sum. Whether $d_1$ is 1 or 9, it contributes to the sum $S_6$. The parity of $S_6$ changes, but the number of choices for $d_7$ remains constant at 5. The restriction on $d_1$ only affects the total count, not the parity logic.

Q: Why is the number of choices for $d_7$ always 5 regardless of the previous sum?

Think of the set of digits $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. There are five even digits and five odd digits. If the sum of the first six digits is even, you need an even digit to keep the total sum even. If the sum is odd, you need an odd digit to make the total sum even. In both cases, you are choosing from a set of exactly five available digits.

Question

If the number of seven-digit numbers, such that the sum of their digits is even, is $m \cdot n \cdot 10^n$; $m, n \in \{1, 2, 3, \dots, 9\}$, then $m + n$ is equal to _______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a grand, seven-digit combination lock. You have seven empty slots, $d_1, d_2, d_3, d_4, d_5, d_6, d_7$, waiting to be filled. The challenge is that the sum of these digits must be even.

This is not a problem of brute force; it is a problem of symmetry and balance. We begin with the first slot, $d_1$. As we know, a seven-digit number cannot start with zero.

Thus, $d_1$ has exactly $9$ choices: $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. For the next five slots, $d_2$ through $d_6$, the rules are relaxed. Each can be any digit from $0$ to $9$, giving us $10$ choices for each of these five positions.

By the fundamental principle of counting, the number of ways to fill these first six slots is:

$$9 	imes 10^5$$

### The Parity Principle

Now, we arrive at the seventh slot, $d_7$. This is where the beauty of the problem unfolds. We do not need to calculate the sum of the first six digits, $S_6 = d_1 + d_2 + d_3 + d_4 + d_5 + d_6$. We only care about its parity—whether it is even or odd.

If $S_6$ is even, we need $d_7$ to be even to ensure the total sum is even. The even digits are $\{0, 2, 4, 6, 8\}$, which gives us exactly $5$ choices.

If $S_6$ is odd, we need $d_7$ to be odd to make the total sum even. The odd digits are $\{1, 3, 5, 7, 9\}$, which also gives us exactly $5$ choices. Regardless of the sum of the first six digits, the seventh digit always has exactly $5$ choices to satisfy the condition.

### The Final Calculation

With this insight, the total number of valid seven-digit numbers is simply the product of the choices for each slot:

$$9 	imes 10^5 	imes 5 = 45 	imes 10^5$$

The problem asks us to match this to the form $m \cdot n \cdot 10^n$. We can rewrite $45$ as $9 	imes 5$. Thus, our expression becomes:

$$9 	imes 5 	imes 10^5$$

Comparing this to $m \cdot n \cdot 10^n$, we identify $m = 9$ and $n = 5$. The final step is to find $m + n$, which is:

$$9 + 5 = \mathbf{14}$$

We have navigated the complexity of the problem by finding the underlying parity structure, turning a potentially tedious counting exercise into a moment of mathematical clarity. Whenever you see a parity constraint, look for the balancing act.

Animated Solution for Mathematics - Permutations and Combinations: If the number of seven-digit numbers, such that the sum of their digits is even, is $m \cdot n \cdot 1 0^{n}$ ; $m, n \in {1, 2, 3, \dots, 9}$ , then $m + n$ is equal to _______

Enter Numerical Value:

Visualized Solution

Structure of a $7$ -Digit Number

Constraint on the First Digit

Constraints on Middle Digits

The Parity Principle for $S_{6}$

Analyzing the Last Digit (Even Case)

Analyzing the Last Digit (Odd Case)

Conclusion on the Last Digit

Calculating Total Valid Numbers

Simplifying the Expression

Matching with $m \cdot n \cdot 1 0^{n}$

Finding $m$ and $n$

Final Result: $m + n$

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

The Parity Principle

The Final Calculation

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