<Question>
If $n \equiv 2 \pmod{7}$, then find the remainder when $(n + 2)(n + 4)(n + 6)$ is divided by 7.
</Question>

<think>
<Plan_1>
Okay, so I need to find the remainder when (n + 2)(n + 4)(n + 6) is divided by 7, given that n ≡ 2 mod 7. Let me start by understanding what the problem is asking. Since n is congruent to 2 modulo 7, that means when n is divided by 7, the remainder is 2. So n can be written as 7k + 2 for some integer k. But maybe I don't need to substitute that in right away. Let me think step by step.

First, since n ≡ 2 mod 7, maybe I can substitute n with 2 in the expression (n + 2)(n + 4)(n + 6) modulo 7.

<prompt_1.1> Substitute n with 2 in the expression (n + 2)(n + 4)(n + 6) modulo 7 and calculate the result. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Because if two numbers are congruent modulo 7, then their expressions involving additions and multiplications will also be congruent modulo 7. That seems like a useful property here. Let me verify that.

Yes, if a ≡ b mod m and c ≡ d mod m, then a + c ≡ b + d mod m and a*c ≡ b*d mod m. So, applying this to each term in the product:

(n + 2) ≡ (2 + 2) mod 7 = 4 mod 7

(n + 4) ≡ (2 + 4) mod 7 = 6 mod 7

(n + 6) ≡ (2 + 6) mod 7 = 8 mod 7 ≡ 1 mod 7 (since 8 - 7 = 1)

So then the product is 4 * 6 * 1 mod 7. Let me compute that.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
4 * 6 = 24, and 24 mod 7. Hmm, 7*3=21, so 24 - 21 = 3. So 24 mod 7 is 3. Then 3 * 1 = 3. So the product is 3 mod 7.
-----

But let me check with another example to make sure. Let's pick n = 9, which is 7 + 2, so 9 ≡ 2 mod 7.

<prompt_2.1> Calculate (n + 2)(n + 4)(n + 6) for n = 9 and find the remainder when divided by 7. </prompt_2.1>

Another example: n = 16, which is 14 + 2, so 16 ≡ 2 mod 7.

<prompt_2.2> Calculate (n + 2)(n + 4)(n + 6) for n = 16 and find the remainder when divided by 7. </prompt_2.2>

Wait, but let me try n = 2, which is the smallest positive integer ≡2 mod7. Then (2+2)(2+4)(2+6) = 4*6*8.
4*6=24, 24*8=192. 192 divided by 7: 7*27=189, so remainder 3. Yep, same result.

Alternatively, maybe there's a smarter way using algebra. Let's see.

<prompt_2.3> Explore an alternative algebraic method to solve the problem and confirm the remainder. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
Then compute (9 + 2)(9 + 4)(9 + 6) = 11 * 13 * 15. Let's calculate that:

11 * 13 = 143. Then 143 * 15. Let's do 143 * 10 = 1430, 143 *5 = 715, so total is 1430 + 715 = 2145. Now divide 2145 by 7. Let's see: 7*300 = 2100. 2145 - 2100 = 45. 45 divided by 7 is 6 with remainder 3. So yes, remainder 3.
</execution_2.1>

<execution_2.2>
Then (16 + 2)(16 + 4)(16 + 6) = 18*20*22. Let's compute that: 18*20 = 360, 360*22 = 7920. Now divide 7920 by 7. Let's see, 7*1131 = 7917, so 7920 - 7917 = 3. Remainder 3 again.
</execution_2.2>

<execution_2.3>
If n ≡2 mod7, then let's consider the three terms:

n + 2 ≡2 +2=4 mod7

n +4≡2 +4=6 mod7

n +6≡2 +6=8≡1 mod7

So the product is 4*6*1=24≡3 mod7. Yep, same answer.

Alternatively, maybe factor something out? Let's see.

Alternatively, note that (n + 2)(n +4)(n +6) can be written as (n +2)(n +4)(n +6). Let me see if there's a pattern here. The terms are n +2, n +4, n +6, which are consecutive even numbers if n is even, but modulo7, they are 4,6,1 as we saw. Alternatively, perhaps this product can be related to n^3 + something. Let me multiply it out.

Multiply (n +2)(n +4) first: n^2 +6n +8. Then multiply by (n +6):

(n^2 +6n +8)(n +6) = n^3 +6n^2 +6n^2 +36n +8n +48 = n^3 +12n^2 +44n +48.

But modulo7, coefficients can be reduced:

12n^2 mod7: 12-7=5, so 5n^2

44n mod7: 44 divided by7 is 6*7=42, so 44-42=2, so 2n

48 mod7: 48 - 6*7=48-42=6

So the expression modulo7 is n^3 +5n^2 +2n +6.

But since n ≡2 mod7, substitute n=2:

2^3 +5*(2)^2 +2*2 +6 =8 +5*4 +4 +6=8 +20 +4 +6=38. 38 mod7: 7*5=35, 38-35=3. So 3 mod7. Same result. So even when expanding the polynomial and reducing coefficients modulo7, we still get 3.

Alternatively, perhaps using properties of modular arithmetic. Since modulus is prime (7 is prime), maybe Fermat's little theorem applies. Fermat's little theorem says that n^6 ≡1 mod7 if n not divisible by7. But here, we have a cubic expression. Maybe not directly applicable, but let's see.

Alternatively, note that (n +2)(n +4)(n +6) can be written as [(n +2)(n +4)(n +6)]. Let me see if there's a way to relate this to a factorial or something. But modulo7, 6≡-1, 4≡-3, 2≡2. Wait, maybe not.

Alternatively, since n ≡2 mod7, then n +2 ≡4, n +4≡6, n +6≡1 as before. So product is 4*6*1=24≡3 mod7. So regardless of the approach, the answer is 3.
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2, execution_2.3:
I think that's solid. All different approaches lead to the same conclusion.
So the remainder is 3.

**Final Answer**
The remainder when $(n + 2)(n + 4)(n + 6)$ is divided by 7 is \boxed{3}.
</Plan_3>

<Execution_3>
No prompt to execute.
</Execution_3>

<Final_answer>
Therefore, the remainder when $(n + 2)(n + 4)(n + 6)$ is divided by 7 is $\boxed{3}$.
</Final_answer>
</think>