How many 6 digit numbers are possible where the sum of the 1st and last digit is 7 and the sum of all the digits is 35?

We are tasked with finding how many 6-digit numbers exist where the sum of the 1st and last digit is 7 and the sum of all digits is 35.

Let the digits of the number be denoted as:

\[
a_1, a_2, a_3, a_4, a_5, a_6
\]

where \(a_1\) is the first digit, and \(a_6\) is the last digit. We are given the following conditions:

1. \( a_1 + a_6 = 7 \)
2. The sum of all digits is 35: \( a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 35 \).

Using condition (1), we can express \(a_6 = 7 - a_1\). Substituting this into condition (2):

\[
a_1 + a_2 + a_3 + a_4 + a_5 + (7 - a_1) = 35
\]

Simplifying:

\[
a_2 + a_3 + a_4 + a_5 + 7 = 35
\]

\[
a_2 + a_3 + a_4 + a_5 = 28
\]

Thus, the sum of the middle digits \(a_2, a_3, a_4, a_5\) must equal 28, and each of these digits can take values between 0 and 9.

Now, let's analyze how many solutions exist for \(a_2 + a_3 + a_4 + a_5 = 28\). We can treat this as a problem of finding the number of non-negative integer solutions to the equation where each digit is constrained to be between 0 and 9.

We need to count all valid combinations of \(a_2, a_3, a_4, a_5\) such that:

\[
0 \leq a_2, a_3, a_4, a_5 \leq 9 \quad \text{and} \quad a_2 + a_3 + a_4 + a_5 = 28
\]

### Step 1: Without restrictions, the number of non-negative integer solutions is the same as the stars and bars formula, which gives solutions for sums. However, because of the upper bound restrictions \(0 \leq a_2, a_3, a_4, a_5 \leq 9\), we will need to use inclusion-exclusion to handle cases where some digits exceed 9.

I'll compute the number of valid solutions.from sympy import symbols, Eq, solve

# Variables for digits a2, a3, a4, a5
a2, a3, a4, a5 = symbols('a2 a3 a4 a5', integer=True)

# Equation for the sum of the digits
eq = Eq(a2 + a3 + a4 + a5, 28)

# Finding all non-negative integer solutions where each digit is between 0 and 9
solutions = 0
for i in range(10):
    for j in range(10):
        for k in range(10):
            l = 28 - i - j - k
            if 0 <= l < 10:
                solutions += 1

solutionsThere are 165 valid combinations of the digits \(a_2, a_3, a_4, a_5\) such that their sum equals 28 and each digit is between 0 and 9.

### Step 2: Accounting for the first and last digit:
- Since \( a_1 + a_6 = 7 \), the valid pairs \((a_1, a_6)\) are:
  \[
  (0,7), (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (7,0)
  \]
  This gives 8 possible pairs for \(a_1\) and \(a_6\).

### Final Answer:
The total number of 6-digit numbers that satisfy the given conditions is:

\[
8 \times 165 = 1320
\]

Thus, there are **1320** possible 6-digit numbers.

Would you like more details on any part of this process? Here are five related questions you might find interesting:

1. How does the stars and bars method work in combinatorics?
2. How is inclusion-exclusion applied to problems with upper bounds?
3. Can you explain the role of constraints in combinatorial problems?
4. What other methods could be used to solve this type of digit sum problem?
5. How would this problem change if the digits had to follow a different rule?

**Tip:** When solving constrained combinatorics problems, always check if inclusion-exclusion or generating functions might be needed to handle complex cases.

Discover how to count 6-digit numbers where the sum of the first and last digit is 7, and the total sum of all digits is 35. This combinatorial problem involves constraints and solutions using mathematical theorems like Stars and Bars and Inclusion-Exclusion. Suitable for advanced high school students.

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