3 times of the last digit of  117627+301^501

To find "3 times the last digit" of \( 117627 + 301^{501} \), we first need to determine the last digit of the sum. Let's break this down step by step:

### Step 1: Find the last digit of \( 117627 \)
The last digit of \( 117627 \) is \( 7 \).

### Step 2: Find the last digit of \( 301^{501} \)
To find the last digit of \( 301^{501} \), we only need to consider the last digit of \( 301 \), which is \( 1 \). 

Any power of a number ending in \( 1 \) will always have a last digit of \( 1 \). Therefore, the last digit of \( 301^{501} \) is \( 1 \).

### Step 3: Find the last digit of \( 117627 + 301^{501} \)
Now, add the last digits from the previous steps:
\[ 7 + 1 = 8 \]
So, the last digit of \( 117627 + 301^{501} \) is \( 8 \).

### Step 4: Calculate 3 times the last digit
Finally, multiply the last digit by 3:
\[ 3 \times 8 = 24 \]

The last digit of \( 24 \) is \( 4 \).

### Final Answer:
3 times the last digit of \( 117627 + 301^{501} \) is \( \boxed{4} \).

Would you like any further details or have any questions?

Here are some related questions you might find interesting:
1. What is the significance of the last digit in modular arithmetic?
2. How do powers of numbers affect the last digit?
3. Can you find the last digit of \( 12345^{6789} \)?
4. What happens when you add large numbers—how can you simplify the last digit calculation?
5. How would you find the last two digits of a large sum?
6. What are cyclic numbers, and how do they relate to problems like this?
7. How can we generalize this problem to any base, not just base 10?
8. Can you calculate the last digit of \( 2^{1000} \)?

**Tip:** Understanding modular arithmetic is crucial when dealing with large exponents or sums to quickly determine last digits or remainders.

Learn how to calculate 3 times the last digit of 117627 + 301^501 with a detailed step-by-step solution. This problem involves modular arithmetic and exponentiation concepts. Suitable for students in Grades 9-12.

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