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

Let's examine the set of numbers $\{a,a+2,a+4\}$ where $a$ must be a natural number:Among any three consecutive odd numbers, exactly one must be a multiple of $3$.For all three numbers to be prime simultaneously, the one that is a multiple of $3$ must equal $3$ itself (since $3$ is the only prime multiple of $3$).Let's test this condition:If $a=3$:$a=3$ (Prime)$a+2=3+2=5$ (Prime)$a+4=3+4=7$ (Prime)This gives us the valid prime triplet $\{3,5,7\}$.If we pick any other value for $a$, one of the numbers in the triplet will be a composite multiple of $3$ (for example, if $a=5$, the triplet is $\{5,7,9\}$, where $9$ is composite).Thus, $a=3$ is the only valid solution, meaning there is exactly one possible value for $a$.