Prove multiplication closure for the sequence of 1,4,7,10.... I am reading the following problem:

If S = ${1, 4, 7, 10, 13, 16, 19, ...}$ and $a \in S\space$ and $b\in
 S \space$ then if $a = b\cdot c\space$ prove that $c \in S$

My approach:
The elements of $S$ are of the form $1 + 3n\space$ so $a = 1 + 3\cdot x\space$ and $b = 1 + 3 \cdot y\space$ and $x = y + j\space$ for some $j >= 1$
If $a = bc \implies c \mid a \space \And \space b\mid a$
Now $b \mid a \implies (1 + 3y)\mid (1 + 3x) \implies (1 + 3y) \mid (1 + 3(y + j)) \implies (1 + 3y) | (1 + 3y + 3j) \implies (1 + 3y) | (1 + 3y) + 3j$
Since $1 + 3y \mid (1 + 3y) \implies (1 + 3y) \mid 3j$ and we keep that.
Now we know that
$$c = \frac{a}{b} = \frac{1 + 3x}{1 + 3y} = \frac{1+3(y + j)}{1 + 3y} = \frac{1 + 3y + 3j}{1 + 3y} = \frac{1 + 3y}{1 + 3y} + \frac{3j}{1+3y} = 1 + \frac{3j}{1 + 3y}$$
But we have shown that $1 + 3y \mid 3j \implies 3j = k(1 + 3y)\space$ where $1 + 3y$ is a positive integer.
Hence $$c = 1 + \frac{k(1 + 3y)}{1 + 3y} = 1 + k$$
But now I am stuck because I need to show that $k$ is a multiple of $3$ to finish the proof and I am not sure how to do that.
I know that $$k = 3\frac{j}{1 + 3y}$$ but the $\frac{j}{1 + 3y}$ is not an integer I think so I a messing up somewhere.
Update:
Following the comment of @Infinity_hunter
Let $c = r + 3k\space$ for some $r >= 1$. We have:
$$c = \frac{a}{b} = \frac{1 + 3x}{1 + 3y} = 1 + \frac{3j}{1 + 3y}$$
This was shown earlier.
But
$$c = r + 3k = 1 + \frac{3j}{1 + 3y} $$
We know that  $$\frac{3j}{1 + 3y}$$ since we have shown that $1 + 3y \mid 3j$ so let $\frac{3j}{1 + 3y} = p$
So we have:
$c = r + 3k = 1 + p \implies c = r - 1 - p + 3k \implies c = (r - 1 - p) + 3k$
And I am not sure how to progress from here.
 A: Straightforward multiplication affords the easiest visualization. Any one member of the set which is formed by the given sequence can be represented by $3a+1$, and any second member of the set can be represented by $3b+1$. Multiplying:
$$(3a+1)(3b+1)=9ab+3b+3a+1=3(3ab+b+a)+1$$
We can readily let $c=3ab+b+a$, from which the product is $3c+1$, and hence a member of the set. This demonstrates closure under multiplication.
Added in response to comment. I originally answered the question posed in the title. The slightly distinct internal question (involving numbers $a,b,c$) can be addressed by the same basic method. Let $a=3A+1, b=3B+1, c=3C+r$ where $r$ can be $0,1,2$, which covers all possibilities.
$$3A+1=(3B+1)(3C+r)=9BC+3Br+3C+r=3(3BC+Br+c)+r$$
Plainly $3$ times something ($A$) plus $1$ equals $3$ times something else ($3BC+Br+c$) plus $r$ can only hold if $r=1$, and specifically not the other possibilities $r=0,2$. Thus $c=(3C+1) \in S$
A: Since $a,b\in S$ we have $a=3x+1$ and $b=3y+1$ for some integers $x,y$. So $$3x+1 = (3y+1)c$$ and thus  $$c = 3\underbrace{(x-yc)}_k +1 = 3k+1 \implies c\in S$$
A: Take $c = r + 3k_1, a = 1 + 3k_2$ and $b = 1 + 3k_3$ where$k_1,k_2,k_3 $ and $r$ are integers. By Division theorem we can chose $r$ such that $r =0,1 $ or $2$. Now we know that $a = bc$ so
$$3k_2 + 1 = (r+3k_1)(1+3k_3) = r + 3k_3r+ 3k_1 + 9k_1k_3$$
So by rearranging terms we have
$$ 1 - r = 3( k_3r + k_1+3k_1k_3 - k_2)$$ which implies that $3$ divides only $1-r$ which is possible only when $r=1$.
