To find "3 times the last digit" of \( 117^{27} + 301^{501} \), we need to determine the last digit of the sum first. Let's go through the steps:

### Step 1: Find the last digit of \( 117^{27} \)
The last digit of \( 117 \) is \( 7 \). Now, we need to find the last digit of \( 7^{27} \).

The powers of 7 cycle every 4 in their last digit:
- \( 7^1 = 7 \) (last digit is 7)
- \( 7^2 = 49 \) (last digit is 9)
- \( 7^3 = 343 \) (last digit is 3)
- \( 7^4 = 2401 \) (last digit is 1)

So, the pattern of last digits for \( 7^n \) is: 7, 9, 3, 1, and this cycle repeats every 4 terms.

Since \( 27 \mod 4 = 3 \), the last digit of \( 7^{27} \) is the same as the last digit of \( 7^3 \), which is \( 3 \).

### Step 2: Find the last digit of \( 301^{501} \)
The last digit of \( 301 \) is \( 1 \). Any power of a number ending in \( 1 \) has a last digit of \( 1 \). So, the last digit of \( 301^{501} \) is \( 1 \).

### Step 3: Find the last digit of \( 117^{27} + 301^{501} \)
Now, add the last digits from the previous steps:
\[ 3 + 1 = 4 \]
So, the last digit of \( 117^{27} + 301^{501} \) is \( 4 \).

### Step 4: Calculate 3 times the last digit
Finally, multiply the last digit by 3:
\[ 3 \times 4 = 12 \]

The last digit of \( 12 \) is \( 2 \).

### Final Answer:
3 times the last digit of \( 117^{27} + 301^{501} \) is \( \boxed{2} \).

Would you like any further details or have any questions?

Here are some related questions you might find interesting:
1. How do you determine the last digit of a number raised to a large power?
2. What is the pattern in the last digits of powers of any single-digit number?
3. How does the cycle of last digits help in simplifying large exponent problems?
4. What are other examples of numbers with simple cyclic patterns?
5. How do you calculate the last digit of sums involving large numbers?
6. Can this method be extended to find the last two digits of a number?
7. What is the importance of modulo operations in problems involving last digits?
8. How would the problem change if the base of the exponent were different, say, 9 or 8?

**Tip:** Recognizing the cyclical nature of last digits in powers of integers can greatly simplify complex calculations.

Learn how to find 3 times the last digit of 117^27 + 301^501 using modular arithmetic and exponentiation rules. Understand the cyclic nature of last digits in powers of integers to simplify large exponent problems. Suitable for high school and above.

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