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!
 A: 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 $d(5k) = 4$.
A: The divisors of 5k are $\{1, 5, k, 5k\}$ if and only if k is a prime number or if $k = 25$.
The divisors of $5 \times 25$ are $\{1, 5, 25, 125\}$. Their sum is $3 \pmod 9$.
Otherwise, you want to compute $1 + 5 + k + 5k \pmod 9 \equiv 6(1+k) \pmod 9$ for all prime numbers $k$ (Except $k=5$).
For $k = 0 \text{ to } 8 \pmod 9$, the values of $6(1+k) \pmod 9$ are $6, 3, 0, 6, 3, 0, 6, 3$.
If $k$ is a multiple of $3$, then $k \in \{3, 6, 9, 12, \dots\}$.
Except for $k = 3$, none of those numbers are prime numbers.
This means that, except for $k = 3$, k cannot be congruent to $0, 3$, or $6$ modulo $9$.
This eliminates the bolded numbers in the sequence of possible values
6, $3$, $0$, 6, $3$, $0$, 6, $3$.
The conclusion is


*

*$d(5k) = 4$ if and only if k = 25 or if k is a prime number other than 5.

*For $k = 3$ the sum of the divisors of $5 \times 3$ is $6$ modulo $9$.

*For $k = 25$ the sum of the divisors of $5 \times 25$ is $3$ modulo
$9$.

*Otherwise, k cannot be congruent to $0, 3$, or $6$ modulo $9$.

*For $k \equiv 1, 4$, or $7 \pmod 9$, the sum of the divisors of $5k$ is
$6$ modulo $9$.

*For $k \equiv 2, 5$, or $8 \pmod 9$, the sum of the divisors of 5k is
$3$ modulo $9$.

A: After reading the answers, I think I have found my own way to the solution. It is based on the property of the prime numbers, for $p\gt 3$ it belongs to Case 1: $1+6x$ or Case 2: $5+6x$. So replacing:
Case 1:
$1+5+p+5p = 1+5+(1+6x)+5(1+6x)=1+5+1+6x+5+30x = 12 + 36x = 12 (1 + 3x)$
And
$12(1+3x)\ mod\ 9 = 3 * (1 + (3\ or \ 6\ or\ 0\ (mod 9))$ so
$3(1+3)\ or\ 3(1+6)\ or\ 3(1+0)$ is respectively $12, 21, 3$ and $(mod\ 9)$ respectively $3,3,3$, so when $p=1+6x$ then $1+5+p+5p \equiv 3\ (mod\ 9)$ 
Case 2:
$1+5+p+5p = 1+5+(5+6x)+5(5+6x)=1+5+5+6x+25+30x = 36 + 36x = 36 (1 + x)$
And
$36(1+x)\ mod\ 9 = 0 * ((1+x)\ mod\ 9)$ so when $p=5+6x$ then $1+5+p+5p \equiv 0\ (mod\ 9)$ 
