Basic question about modular arithmetic applied to the divisor sum function $\sigma(n)$ when $n=5p$ While studying the divisor sum function $\sigma(n)$ (as the sum of the divisors of a number) I observed that the following expression seems to be true always (1):

$\forall\ n=5p, p\in\Bbb P,\ p\gt 5,\ if\ d(5p)=4\ then\ \sigma(5p)=3,0\ (mod\ 9)$

Meaning that if the number $n=5k$ has only four divisors, so $5$ and other prime $p$ divide $n$, then the sum of the divisors is congruent to 3 or 0 modulo 9. I just observed that manually in the range $[1..400]$.
The divisors of a $5p$ number as the type above are $\{1,5,p,5p\}$
E.g.
$d(35) = \{1,5,7,35\}$
$d(55) = \{1,5,11,55\}$
And the sum of the divisors is $\sigma(5p)= 1+5+p+5p$, thus (2):
$(1+5+p+5p)\ mod\ 9 = 1 + 5 + ((p + 5p)\ mod\ 9) =$
$6 + ((p + 5p)\ mod\ 9) = 6 + (6p\ mod\ 9)$
... but from that point I am not sure how to conclude that for the case above (1) it will be $3$ or $0$.
I think that this must happen at (2) if (1) is true:
$6p\ mod\ 9 = 6 * (p\ mod\ 9) \equiv 3, 6\ (mod\ 9)$
and then is true only if:
$p\equiv 1,x?\ (mod\ 9)$
Any help or hint is very appreciated, or a counterexample if (1) is wrong and then the congruence is false, thank you!

Note that if $k=3$, then $\sigma(15) =1+3+5+15 = 24$ is congruent to $6$ modulo $9$.  This is the only counterexample, as we shall show.
Note that the equation $6k \equiv 3, 6 \pmod{9}$ is equivalent to saying that $3$ divides $6k$, but $3^2$ does not.  Clearly the former is true, and the latter is true for all $k \ne 3$, since $k$ must be either a prime or equal to $25$ in order for $d(5k) = 4$.
