If a, a + 2, and a + 4 are all prime numbers, how many possible values can a…

Concept: Among any three integers spaced 2 apart — a, a+2, a+4 — exactly one of the three is always divisible by 3, regardless of which residue class a falls into modulo 3. For all three numbers to be prime simultaneously, that multiple of 3 must itself be the prime 3, since no other multiple of 3 can be prime.

Application:

1. If a is divisible by 3 (remainder 0 mod 3), then a itself must equal 3 to be prime. This gives 3, 5, 7 — all three are prime, so this case is valid.

2. If a leaves remainder 1 mod 3, then a+2 is the multiple of 3, so a+2 must equal 3, giving a = 1. But 1 is not a prime number, so this case is invalid.

3. If a leaves remainder 2 mod 3, then a+4 is the multiple of 3, so a+4 must equal 3, giving a = -1. Negative numbers are not prime, so this case is invalid too.

Cross-check: Testing a few other prime starting values confirms the modular argument: a = 5 gives 5, 7, 9; a = 7 gives 7, 9, 11; and a = 11 gives 11, 13, 15. In each case, the member divisible by 3 is greater than 3 and therefore composite.

Therefore a has exactly one admissible value, namely a = 3. The number 3 is the value of a; it is not the number of possible values. Hence the required count is One.