Jay notices that there are $n$ primes that form an arithmetic sequence with common difference $12$. What is the maximum possible value for $n$?

[i]Proposed by James Lin[/i]
To determine the maximum number of primes \( n \) that can form an arithmetic sequence with a common difference of 12, we need to consider the properties of prime numbers and modular arithmetic.

1. **Prime Number Forms**:
   Prime numbers greater than 3 can be expressed in the form \( 6k \pm 1 \) because any integer can be written as \( 6k, 6k+1, 6k+2, 6k+3, 6k+4, \) or \( 6k+5 \). Among these, \( 6k, 6k+2, 6k+3, \) and \( 6k+4 \) are divisible by 2 or 3, and thus cannot be prime (except for 2 and 3 themselves).

2. **Arithmetic Sequence Modulo 5**:
   Since the common difference is 12, we consider the sequence modulo 5. Note that \( 12 \equiv 2 \pmod{5} \). Therefore, the sequence modulo 5 will have a common difference of 2.

3. **Sequence Analysis**:
   Let's analyze the sequence modulo 5:
   \[
   a, a+12, a+24, a+36, \ldots \equiv a, a+2, a+4, a+6, \ldots \pmod{5}
   \]
   This sequence will cycle through all residues modulo 5. Specifically:
   \[
   a, a+2, a+4, a+6 \equiv a, a+2, a+4, a+1, a+3 \pmod{5}
   \]
   After 5 terms, the sequence will include all residues modulo 5.

4. **Prime Consideration**:
   Since the sequence will include all residues modulo 5, one of these residues will be 0 modulo 5. This means that one of the terms in the sequence will be divisible by 5. The only prime number that is divisible by 5 is 5 itself.

5. **Maximum Sequence Length**:
   Therefore, the maximum number of primes in such a sequence before encountering a non-prime (or a composite number) is 4, plus the prime number 5 itself. This gives us a total of 5 primes.

Thus, the maximum possible value for \( n \) is:
\[
\boxed{5}
\]