<Question>
Let $n$ be a positive integer.  What is the greatest possible value of $\gcd(n + 7, 2n + 1)$?
</Question>

<think>
<Plan_1>
Okay, so I need to find the greatest possible value of the greatest common divisor (gcd) of (n + 7) and (2n + 1), where n is a positive integer. Hmm, let me think about how to approach this. I remember that gcd(a, b) is the largest number that divides both a and b. So, I need to find the maximum value such that this number divides both n + 7 and 2n + 1.

Let me recall some properties of gcd. One method to find gcd of two numbers is the Euclidean algorithm, which involves subtracting multiples of the smaller number from the larger one until we reach the gcd. Maybe I can apply that here.

Let me denote a = n + 7 and b = 2n + 1. Then, gcd(a, b) = gcd(n + 7, 2n + 1). Using the Euclidean algorithm, I can subtract 2 times (n + 7) from (2n + 1) to eliminate n. Let's compute that:

<prompt_1.1> Apply the Euclidean algorithm to find gcd(n + 7, 2n + 1) by eliminating n. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
2n + 1 - 2*(n + 7) = 2n + 1 - 2n - 14 = -13.

Wait, that's interesting. So, gcd(n + 7, 2n + 1) = gcd(n + 7, 13). Because after subtracting 2*(n + 7) from 2n + 1, we get -13, and gcd is always non-negative, so it's 13. Therefore, the gcd of (n + 7) and (2n + 1) is the same as the gcd of (n + 7) and 13.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Since 13 is a prime number, the possible gcd values are 1 and 13. Therefore, the greatest possible value of the gcd is 13, provided that 13 divides (n + 7). If 13 divides (n + 7), then n + 7 is a multiple of 13, which means n ≡ -7 mod 13.
But since n is a positive integer, we can choose n such that n = 13k - 7 for some integer k ≥ 1. Let's check if this works.

For example, let's take k = 1: n = 13*1 - 7 = 6.

<prompt_2.1> Verify that gcd(6 + 7, 2*6 + 1) is 13. </prompt_2.1>

If k = 2: n = 13*2 -7 = 19.

<prompt_2.2> Verify that gcd(19 + 7, 2*19 + 1) is 13. </prompt_2.2>

But wait, let me verify this. If n +7 is divisible by 13, then 2n +1 would be 2*(13m -7) +1 = 26m -14 +1 = 26m -13, which is 13*(2m -1), so indeed divisible by 13.
Therefore, both n +7 and 2n +1 are divisible by 13, so gcd is at least 13. Since 13 is prime, the gcd can't be higher than 13.

But hold on, let me check with another example. Suppose n = 6, as before.

<prompt_2.3> Verify the gcd for n = 6, n = 19, and n = 13k - 7 for k = 3. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
Then, n + 7 = 13, which is divisible by 13. Let's compute 2n + 1 = 12 + 1 = 13. So, gcd(13, 13) = 13.
</execution_2.1>

<execution_2.2>
Let's check gcd(19 +7, 2*19 +1) = gcd(26, 39). Both 26 and 39 are divisible by 13, so gcd is 13, which is correct. Similarly, for n =6, gcd(13,13) =13.
</execution_2.2>

<execution_2.3>
Then, n +7 = 13, and 2n +1 = 13. So gcd(13,13) =13. Correct. If n =19, then n +7=26, 2n +1=39. gcd(26,39)=13. Correct again. If n =13*3 -7=32, then n +7=39, 2n +1=65. gcd(39,65)=13. Still correct. So, seems like whenever n ≡-7 mod13, which is same as n≡6 mod13, then gcd is13. Otherwise, it's 1.
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2, execution_2.3:
Therefore, the maximum possible gcd is 13.

But let me think again if there's any possibility that the gcd could be higher than 13. Since we reduced the problem to gcd(n +7,13), and 13 is prime, the possible divisors are only 1 and 13. So, the maximum possible is indeed 13, which occurs when 13 divides (n +7).
So, the answer should be 13. Therefore, the greatest possible value is 13.

**Final Answer**
The greatest possible value is \boxed{13}.
</Plan_3>

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

<Final_answer>
Hence, the greatest possible value of $\gcd(n + 7, 2n + 1)$ is $\boxed{13}$.
</Final_answer>
</think>