If a, a + 2, a + 4 are prime numbers, then the number of values of 'a' are:

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 starting values confirms this: a = 5 gives 5, 7, 9 (9 is not prime); a = 7 gives 7, 9, 11 (9 is not prime); a = 11 gives 11, 13, 15 (15 is not prime). Every case beyond a = 3 produces a multiple of 3 greater than 3, which cannot be prime.

Hence exactly one value of a (a = 3) makes a, a+2, and a+4 all prime at once, so the number of values of a is One.