What is the only integer value of $n$ for which $\frac{n+1}{13-n}$ is a positive prime number?
To answer this question, I need to think about what conditions must be satisfied for the expression $\frac{n+1}{13-n}$ to be a positive prime number. First, since the expression is a fraction, I need to make sure that the denominator $13-n$ is not zero, because that would make the fraction undefined. That means $n$ cannot be $13$. Second, since the expression is positive, I need to make sure that both the numerator $n+1$ and the denominator $13-n$ have the same sign. That means either both are positive or both are negative. Third, since the expression is a prime number, I need to make sure that both the numerator $n+1$ and the denominator $13-n$ are coprime, meaning they have no common factors other than $1$. That means they cannot share any prime divisors. Now, I can use these conditions to eliminate some possible values of $n$. For example, if $n$ is greater than $13$, then the denominator $13-n$ is negative, but the numerator $n+1$ is positive, which violates the second condition. So $n$ cannot be greater than $13$. Similarly, if $n$ is less than $-1$, then the numerator $n+1$ is negative, but the denominator $13-n$ is positive, which also violates the second condition. So $n$ cannot be less than $-1$. This means that $n$ must be in the interval $[-1, 13)$. Within this interval, there are only $15$ possible integer values of $n$: $-1, 0, 1, 2, \dots, 12$. I can check each of these values and see if they make the expression $\frac{n+1}{13-n}$ a positive prime number. For $n = 12$, the expression becomes $\frac{13}{1}$, which is $13$, a positive prime number. Therefore, the only integer value of $n$ for which $\frac{n+1}{13-n}$ is a positive prime number is $n = 12$. 