To determine the units digit (the last digit) of the sum $S = 1^{2024} + 2^{2024} + 3^{2024} + \dots + 2024^{2024}$, we need to focus on the units digits of each term in this sum individually, as only the units digit of each term affects the units digit of the final sum.

Step 1: Units Digits Cycles of $n^{2024}$

For any integer $n$, the units digit of $n^{2024}$ depends on the cyclic pattern of the units digit of powers of $n$. Let's go through the typical units digit cycles for numbers 0 through 9.

1. Units digit 0: $0^1, 0^2, \dots$ all end in 0.
2. Units digit 1: $1^1, 1^2, \dots$ all end in 1.
3. Units digit 2: Cycle is $2, 4, 8, 6$ (4 terms).
4. Units digit 3: Cycle is $3, 9, 7, 1$ (4 terms).
5. Units digit 4: Cycle is $4, 6$ (2 terms).
6. Units digit 5: $5^1, 5^2, \dots$ all end in 5.
7. Units digit 6: $6^1, 6^2, \dots$ all end in 6.
8. Units digit 7: Cycle is $7, 9, 3, 1$ (4 terms).
9. Units digit 8: Cycle is $8, 4, 2, 6$ (4 terms).
10. Units digit 9: Cycle is $9, 1$ (2 terms).

Step 2: Determine the Units Digit of Each $n^{2024}$

To find the units digit of $n^{2024}$ for each $n$, we find $2024 \mod \text{cycle length}$ for each relevant units digit cycle:

• For $n$ ending in 0, 1, 5, or 6: The units digit of $n^{2024}$ is the same as $n$, so these terms contribute units digits of 0, 1, 5, and 6, respectively.
• For $n$ ending in 2: The cycle is 4 terms (2, 4, 8, 6). Since $2024 \mod 4 = 0$, we take the last digit in the cycle, which is 6.
• For $n$ ending in 3: The cycle is 4 terms (3, 9, 7, 1). Since $2024 \mod 4 = 0$, we take the last digit in the cycle, which is 1.
• For $n$ ending in 4: The cycle is 2 terms (4, 6). Since $2024 \mod 2 = 0$, we take the last digit in the cycle, which is 6.
• For $n$ ending in 7: The cycle is 4 terms (7, 9, 3, 1). Since $2024 \mod 4 = 0$, we take the last digit in the cycle, which is 1.
• For $n$ ending in 8: The cycle is 4 terms (8, 4, 2, 6). Since $2024 \mod 4 = 0$, we take the last digit in the cycle, which is 6.
• For $n$ ending in 9: The cycle is 2 terms (9, 1). Since $2024 \mod 2 = 0$, we take the last digit in the cycle, which is 1.

Step 3: Sum the Units Digits

Now, we sum the contributions from each possible units digit: $0 + 1 + 6 + 1 + 6 + 5 + 6 + 1 + 6 + 1 = 33$

Step 4: Find the Units Digit of the Total Sum

The units digit of 33 is 3. Therefore, the units digit of $S = 1^{2024} + 2^{2024} + \dots + 2024^{2024}$ is: $\boxed{3}$

Would you like more details on any part of this calculation?

---

1. How do cyclic patterns help in finding powers’ last digits?
2. Can the approach here be used for any power, like $n^{500}$ or $n^{1000}$?
3. How would the solution change if we had a different upper limit, like $1000^{1000}$?
4. What if we had a different modulus (e.g., last two digits)?
5. Why do some numbers (like those ending in 5 or 6) have constant last digits?

Tip: When dealing with powers, finding cyclical patterns for the units digit can greatly simplify large exponent calculations.