Input of a function must give specific output, find number of positive integers that satisfy. Question: The prime factorization of an integer $n \geq 2$ is of the form $({p_1})^{a_1}(p_2)^{a_2} … (p_k)^{a_k}$ where $p_1$, $p_2$, …, $p_k$ are different prime numbers and $a_1, a_2, …, a_k$ are positive integers. Given an input of an integer $n \geq 2$, the Barbeau process outputs the number equal to $n(\frac{a_1}{p_1} + \frac{a_2}{p_2} … + \frac{a_k}{p_k})$. Determine the number of triples (a, b, c) of positive integers such that the Barbeau Process with input $2^a3^b5^c$, outputs $4(2^a3^b5^c)$.
My solution:
First of all, $2, 3$, and $5$ are all primes, which means that we can immediatly use the Barbeau Process.
We know that the input is $2^a3^b5^c$ which means that the output should be equal to $(2^a3^b5^c)(\frac{a}{2} + \frac{b}{3} + \frac{c}{5})$.
So, $(2^a3^b5^c)(\frac{a}{2} + \frac{b}{3} + \frac{c}{5}) = 4(2^a3^b5^c)$.
The $(2^a3^b5^c)$ cancels out on both sides, leaving that $\frac{a}{2} + \frac{b}{3} + \frac{c}{5} = 4$.
This gives that $\frac{15a + 10b + 6c}{30} = 4 \Longrightarrow 15a + 10b + 6c = 120$.
We first realize that since $a, b$, and $c$ are positive integers, and $10b + 6c$ and $120$ are even, $15a$ must also be even.
We also know that $15a < 120$, because otherwise, $b$ and $c$ would not be positive integers.
Testing out the cases of the even values of $a$:
First of all, $a \leq 6$ because of the constraint that $15a < 120$.
$a = 6 \Longrightarrow$ Does not give any values for $b$ and $c$.
$a = 4 \Longrightarrow$ One case, $b = 3, c = 5$.
$a = 2 \Longrightarrow$ Two cases, $b = 6, c = 5$ or $b = 3, c = 10$.
So, there are $\boxed{3}$ total triples $(a, b, c)$; $(4, 3, 5)$, $(2, 6, 5)$, and $(2, 3, 10)$.
Please help me decide if my solution is correct or wrong. Thanks in advance!

It seems like your solution is valid. However, you should state that $a < 8$ instead of $a \leq 6$ because $120/15 = 8$.
