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.