Notice that if $a\mid b$, then, by definition, we have that $b=\ell a$ for some $\ell\in\mathbb{Z}$. Also, if $a\mid c$, then, by definition, we have that $c=\eta a$ for some $\eta\in\mathbb{Z}$. Now use this information effectively:
$$
b\cdot c = (\ell a)\cdot (\eta a) = a(a\eta\ell)=a\gamma,
$$
where $\gamma\in\mathbb{Z}$. Thus, by definition, we can see that $a\mid bc$ if $a\mid b$ and $a\mid c$.
I assume here that you care more about the solution than the process of getting to the solution - the process is just an extremely messy recurrence computation.
I shall solve the set of equations in general by replacing our values with variables and then solving for our particular case by plugging the values back in at the end. We are trying to solve
$$H(n+1)=aH(n)+b$$
$$E(n+1)=dE(n)+eH(n)$$
$$O(n+1)=gO(n)+hE(n)+i$$
with $H(0)=c$, $E(0)=f$, $O(0)=j$
By a WolframAlpha computation we have
$$H(n)=\left(c+\frac{b}{a-1}\right)a^n-\frac{b}{a-1}$$
Hence our second equation becomes
$$E(n+1)=dE(n)+e\left(\left(c+\frac{b}{a-1}\right)a^n-\frac{b}{a-1}\right)$$
Now for convenience we write new constants: $k:=e\left(c+\frac{b}{a-1}\right)$ and $l:=-\frac{eb}{a-1}$, so now the equation reads
$$E(n+1)=dE(n)+ka^n+l$$
By another WolframAlpha computation we have
$$E(n)=\frac{k}{a-d}a^n+\left(f+\frac{k}{d-a}+\frac{l}{d-1}\right)d^n+\frac{l}{1-d}$$
Now again we compactify variables: $m:=\frac{hk}{a-d}$, $o:=hf+\frac{hk}{d-a}+\frac{hl}{d-1}$, $p=\frac{hl}{1-d}+i$, so now the third equation reads
$$O(n+1)=gO(n)+ma^n+od^n+p$$
WolframAlpha wouldn't do the last computation for me, so I used Mathematica to get the solution for $O(n)$ as
$$\frac{-adg^nj+ag^{1+n}j+dg^{1+n}j+adg^{1+n}j-g^{2+n}j-ag^{2+n}j-dg^{2+n}j+g^{3+n}j-a^ndm+a^ngm+a^ndgm-a^ng^2m+dg^nm-g^{1+n}m-dg^{1+n}m+g^{2+n}m-ad^no+d^ngo+ad^ngo-d^ng^2o+ag^no-g^{1+n}o-ag^{1+n}o+g^{2+n}o-adp+agp+dgp-g^2p+adg^np-ag^{1+n}p-dg^{1+n}p+g^{2+n}p}{(g-1)(g-a)(g-d)}$$
(doubtless nicer expressions are possible, but I didn't want to expend the effort to find one)
Plugging your specific variables back in, I get (to three decimal places)
$$H(n)=3227.273(.89)^n+272.727$$
$$E(n)=1420(.89)^n+296.667(.64)^n+83.333$$
$$O(n)=3472.22 (0.144 - 0.128(.64)^n - 910.644(0.88)^n +
3907.02 (0.88)^{1 + n} - 5394(0.88)^{2 + n} + 2400 (.88)^{3 + n} +
14.7226 (0.89)^n)$$
Notice this checks out with your equilibrium solution since $3472.22\times0.144\approx500$ (off by only $3$ ten-thousandths).
Best Answer
A drawing of the function $f(x)=\lceil x\rceil$ can help:
Taking for granted what the drawing says it's very easy to check any of the substitutions (e.g. clearly $\lceil 2.4\rceil=\lceil 2.15\rceil=\lceil 2.836\rceil=3$ simply reading it)
You can too check the truth of the substitutions right into the inequalities. E. g. $\lceil1.7\rceil=2$ satisfies $2-1<1.7\leq2$
Finally, the very name helps: for a number some integer is its ceiling.