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The problem asks for the unit digit of the sum of factorials:

\[
1! + 3! + 5! + 7! + 9! + 11!
\]

Let's calculate the unit digit of each factorial:

1. \(1! = 1\), so the unit digit is **1**.
2. \(3! = 3 \times 2 \times 1 = 6\), so the unit digit is **6**.
3. \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\), so the unit digit is **0**.
4. \(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\), so the unit digit is **0**.
5. \(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880\), so the unit digit is **0**.
6. \(11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 39916800\), so the unit digit is **0**.

Now, we sum the unit digits of these factorials:
\[
1! + 3! + 5! + 7! + 9! + 11! = 1 + 6 + 0 + 0 + 0 + 0 = 7
\]

Thus, the unit digit of the entire sum is **7**.

Would you like more details on any part of the solution?

Here are 5 related questions for practice:
1. What is the unit digit of \(13! + 15! + 17! + 19!\)?
2. How does the factorial of numbers larger than 4 affect the unit digit?
3. What is the unit digit of \(2! + 4! + 6! + 8! + 10!\)?
4. How can you generalize the behavior of factorials for determining unit digits?
5. What is the unit digit of \(100! + 99!\)?

**Tip**: For any number \(n! > 4!\), the factorial will always have a unit digit of zero because the factorial includes the multiplication by 10.

What is the unit digit of 1! + 3! + 5! + 7! + 9! + 11!?

Learn how to find the unit digit of the sum of factorials 1! + 3! + 5! + 7! + 9! + 11!. Explore factorial properties and modular arithmetic, suitable for students in Grades 7-9.

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