Prime triples — If N, N + 2, and N + 4 are all prime numbers, how many possible values of N exist? Explain using divisibility by 3.

Introduction / Context:
This is a classic number theory puzzle about “prime triplets.” Most sets of three odd numbers spaced by 2 cannot all be prime because one is divisible by 3. The task is to determine how many such N exist that make N, N+2, and N+4 all prime simultaneously.

Given Data / Assumptions:

• N is an integer (we are seeking prime values for the three terms).
• N, N+2, N+4 are all required to be prime.
• Primes are integers greater than 1 with exactly two positive divisors.

Concept / Approach:
Consider the three numbers modulo 3. Any three consecutive odd numbers (spaced by 2) will cover all residues mod 3. That means at least one of them must be divisible by 3. The only way to avoid a composite member is if that divisible-by-3 term is actually 3 itself, which is prime. This restriction pins down a unique feasible value for N.

Step-by-Step Solution:
Among N, N+2, N+4, one is ≡ 0 (mod 3).If that term is greater than 3, it becomes composite and the triple fails.Thus, the only viable case is when the 0 (mod 3) term equals 3 itself.Set N = 3 to obtain the triple (3, 5, 7), all prime. No other N works.

Verification / Alternative check:
Testing nearby values: N=1 gives (1, 3, 5) with 1 not prime; N=5 gives (5, 7, 9) where 9 is composite; N=11 gives (11, 13, 15) where 15 is composite. Only N=3 produces a fully prime triple.

Why Other Options Are Wrong:

• 2 / 3 / None of these / 4: These overcount; modular reasoning confines the possibility to a single case.

Common Pitfalls:
Forgetting the role of 3 in modular cycles; counting 1 as a prime; assuming multiple solutions without checking divisibility by 3.

Final Answer:
1