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The problem you provided has two parts:

1. **Part 1**: 
   The expression is \( (987654)^{nnn} = \ldots n \). This likely means that the result of raising 987654 to some power related to \( nnn \) ends in \( n \). The challenge here is to figure out what \( n \) is based on this information.

2. **Part 2**: 
   The problem asks to "Halle la última cifra de \( (n + 3)^{(n + 2)} \)," which means "find the last digit of \( (n + 3)^{(n + 2)} \)."

### Solution Approach:

#### Part 1:
To solve the equation \( (987654)^{nnn} \equiv n \pmod{10} \), we need to analyze the last digit of \( 987654 \) raised to various powers.

- The last digit of 987654 is 4. So, we are interested in the powers of 4 modulo 10.
- The powers of 4 repeat every 2 cycles when considering the last digits:
  - \( 4^1 = 4 \) (last digit is 4),
  - \( 4^2 = 16 \) (last digit is 6),
  - \( 4^3 = 64 \) (last digit is 4),
  - \( 4^4 = 256 \) (last digit is 6), and so on.

Thus, the last digit of \( (987654)^{nnn} \) will depend on whether \( nnn \) is odd or even:
- If \( nnn \) is odd, the last digit is 4.
- If \( nnn \) is even, the last digit is 6.

Since \( (987654)^{nnn} \equiv n \pmod{10} \), the value of \( n \) must either be 4 (if \( nnn \) is odd) or 6 (if \( nnn \) is even).

#### Part 2:
Now that \( n \) is either 4 or 6, we need to find the last digit of \( (n + 3)^{(n + 2)} \). This gives us two possibilities:
1. If \( n = 4 \), we need the last digit of \( (4 + 3)^{(4 + 2)} = 7^6 \).
2. If \( n = 6 \), we need the last digit of \( (6 + 3)^{(6 + 2)} = 9^8 \).

Let's compute the last digits:

- \( 7^1 = 7 \),
- \( 7^2 = 49 \) (last digit 9),
- \( 7^3 = 343 \) (last digit 3),
- \( 7^4 = 2401 \) (last digit 1), and the cycle repeats every 4 terms.

Thus, \( 7^6 \) has the same last digit as \( 7^2 \), which is 9.

For \( 9^n \), the last digit follows a cycle:
- \( 9^1 = 9 \),
- \( 9^2 = 81 \) (last digit 1),
- \( 9^3 = 729 \) (last digit 9),
- \( 9^4 = 6561 \) (last digit 1), and the cycle repeats every 2 terms.

Thus, \( 9^8 \) has the same last digit as \( 9^2 \), which is 1.

### Conclusion:
- If \( n = 4 \), the last digit of \( (n + 3)^{(n + 2)} \) is 9.
- If \( n = 6 \), the last digit of \( (n + 3)^{(n + 2)} \) is 1.

Do you want more details or clarifications on any step?

Here are 5 related questions for further practice:
1. What is the last digit of \( 12345^{678} \)?
2. Find the last digit of \( 987654^{321} \).
3. How does the cyclical behavior of last digits work for numbers ending in 2?
4. Can you determine the last digit of \( 3^{2023} \)?
5. What is the last digit of \( (n + 5)^{n^2} \) when \( n = 7 \)?

**Tip**: Understanding the cyclical patterns of last digits for powers can simplify many modular arithmetic problems!

Si (987654)^{nnn} = ...n. Halle la última cifra de (n + 3)^{(n + 2)}.

Learn how to find the last digit of expressions like (n + 3)^{(n + 2)} using modular arithmetic. This problem involves cyclical patterns in number theory and is suitable for high school students.

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