To determine if $n^3 + 11$ is divisible by 6 for any integer $n$, we need to check if the expression $n^3 + 11$ is divisible by both 2 and 3.

Step 1: Check for Divisibility by 2

An integer is divisible by 2 if it is even. The term $n^3$ will be:

• Even if $n$ is even (since the cube of an even number is also even).
• Odd if $n$ is odd (since the cube of an odd number is also odd).

Since $11$ is an odd number:

• If $n$ is even, $n^3$ is even, and $n^3 + 11$ will be odd (not divisible by 2).
• If $n$ is odd, $n^3$ is odd, and $n^3 + 11$ will be even (divisible by 2).

Thus, $n^3 + 11$ is divisible by 2 only when $n$ is odd.

Step 2: Check for Divisibility by 3

We analyze $n^3$ modulo 3. There are three possible cases:

1. If $n \equiv 0 \pmod{3}$: Then $n^3 \equiv 0^3 \equiv 0 \pmod{3}$.
2. If $n \equiv 1 \pmod{3}$: Then $n^3 \equiv 1^3 \equiv 1 \pmod{3}$.
3. If $n \equiv 2 \pmod{3}$: Then $n^3 \equiv 2^3 \equiv 8 \equiv 2 \pmod{3}$.

For $n^3 + 11$ to be divisible by 3, $n^3$ must leave a remainder of 1 modulo 3, because $11 \equiv 2 \pmod{3}$ and $1 + 2 \equiv 0 \pmod{3}$. Thus, $n \equiv 1 \pmod{3}$.

Conclusion

For $n^3 + 11$ to be divisible by 6, $n$ must be odd and satisfy $n \equiv 1 \pmod{3}$.

In summary:

• $n$ must be an odd integer.
• $n \equiv 1 \pmod{3}$ (e.g., $n = 1, 7, 13, \ldots$).

Would you like further clarification on this?

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Here are five questions to deepen your understanding:

1. How can we determine if a number is divisible by any composite number, like 6 or 12?
2. What are the rules for determining divisibility by powers of primes, such as 8 or 9?
3. How does modular arithmetic help in proving divisibility properties?
4. Can the same approach be applied to other expressions involving $n$, such as $n^3 + 17$?
5. How does the parity (odd/even nature) of numbers influence divisibility?

Tip: Checking divisibility by composite numbers like 6 or 12 often involves verifying divisibility by their prime factors.