Determining the Last Digit of a Factorial Chain

To find the last digit of a factorial chain like 1! 2! 3! ... 33!, we need to understand how factorials behave with respect to their last digits. We start by calculating the last digits of the factorials up to 4!.

Step 1: Calculate the Last Digits of Factorials Up to 4!

Let's compute the last digits of the factorials up to 4!: 1! 1 last digit is 1 2! 2 last digit is 2 3! 6 last digit is 6 4! 24 last digit is 4

Step 2: Consider Higher Factorials

For n > 5, n! will always end in 0. This is because 5! and higher factorials contain the factor 10, which is 2 times 5. Therefore, we only need to consider the last digits of the factorials from 1! to 4!:

1! 2! 3! 4! 1 2 6 4 33

Step 3: Find the Last Digit of the Factorial Chain to the Power of 33

We need to find the last digit of the entire product raised to the 33rd power: (1! 2! 3! ... 33!)^{33} .

This is equivalent to finding the last digit of 33^{33}. We can analyze the pattern of the last digits of powers of 3: 3^1 3 last digit is 3 3^2 9 last digit is 9 3^3 27 last digit is 7 3^4 81 last digit is 1

The last digits repeat every 4 terms: 3 9 7 1. To determine the correct last digit for 3^{33} , we find the remainder of 33 when divided by 4: 33 mod 4 1. Since the remainder is 1, the last digit of 3^{33} corresponds to the last digit of 3^1 which is 3.

Therefore, the last digit of 1! 2! 3! 4! ... 33!^{33} is boxed{3}.

Alternative Method: Modulo Arithmetic

We can also evaluate this using modulo arithmetic. Let's denote the factorial chain as S. We first determine the residual of the term within the parentheses.

Note that 10 divides the factorials of 10! and above. So we only need to consider 1! 2! ... 8! 9!. All factorials containing the factors 2 and 5 are also divisible by 10, leaving us with:

1! 2! 3! 4! 1 2 6 24 33 Thus, S mod 10 33 mod 10 3.

Now, we consider S^{33} mod 10. We note that:

S^2 9 mod 10 -1 mod 10

S^32 1 mod 10

S^33 S mod 10 3 mod 10

Thus, the last digit of S^{33} is 3.

In conclusion, the last digit of 1! 2! 3! 4! ... 33!^{33} is boxed{3}.