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The question asks for the digit before the zeros at the end of the result of \(20! - 21!\).

We can break it down:

1. \(21! = 21 \times 20!\), so:
   \[
   21! = 21 \times 20!
   \]

2. Therefore, \(20! - 21!\) becomes:
   \[
   20! - 21! = 20! - 21 \times 20! = 20!(1 - 21) = -20! \times 20
   \]

3. Now, the expression simplifies to:
   \[
   -20! \times 20
   \]

4. The digit before the zeros in \(20!\) is what we're concerned with because multiplying it by 20 and taking the negative won't change this. We need to first calculate \(20!\).

   Using a calculator, \(20! = 2,432,902,008,176,640,000\).

5. Multiplying \(20!\) by 20:
   \[
   -20! \times 20 = -48,658,040,163,532,800,000
   \]

6. The number of zeros at the end of this result doesn't change, but we're interested in the digit before the zeros. In the number \(48,658,040,163,532,800,000\), the last non-zero digit is **8**.

So, the correct answer is **D) 8**.

Would you like more details or have any questions? Here are 5 related questions for further practice:

1. What is the number of trailing zeros in \(20!\)?
2. How do we calculate the number of trailing zeros in any factorial \(n!\)?
3. What would \(21! - 22!\) equal to, and what is the last non-zero digit?
4. How can you find the last non-zero digit in a large factorial \(n!\)?
5. What are factorial applications in combinatorics and probability?

**Tip**: Factorials grow very fast, so shortcuts like analyzing the last few digits or using modulo properties can help with large factorial computations.

What is the digit before the zeros at the end of the number equals to 20! – 21!?

Find the digit before the zeros at the end of 20! - 21!. This problem involves factorials, multiplication, and trailing zeros, suitable for advanced high school math students.

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