The conjecture says that for any a, b belong to the the set of non-negative integers ($a$ and $b$ are not necessarily distinct), taking any natural value of $c$; we have always that $$(10^c-1) \cdot \frac{10^a+10^{2a}+1}{3} \cdot \frac{10^b+10^{2b}+1}{3}$$ is palindromic in base $10$. This conjecture was experimented well.


Example: taking $a=2,b=3,c=5$,
we will get $$3367 \cdot 333667 \cdot 99999=112344555443211$$ which is palindromic in base $10$,
my question is to prove or disprove this conjecture.
Note that the conjecture above was proved before editing .
another conjecture that I ask to prove or disprove it is that the sequence of the numbers of the form (10^a+10^2a+1)/3 is the maximally dense sequence in base 10 with the palindromic products property as described above . 



simple conjecture on palindromes in base 10
Using
$$\frac{10^c-1}{9}=\sum\_{m=0}^{c-1} 10^m,$$
the product in question equals
$$\sum\_{n=0}^{2a+2b+c-1}r(n)10^n,$$
where $r(n)$ is the number of times $n$ occurs among the numbers (counted with multiplicity)
$$\begin{matrix}
m+2a+2b,&m+2a+b,&m+2a,\\
m+a+2b,&m+a+b,&m+a,\\
m+2b,&m+b,&m,\\
\end{matrix}$$
for some $m\in\{0,1,\dots,c-1\}$. For a given $n$, each entry in the above $3\times 3$ grid equals $n$ for at most one value of $m$, hence $r(n)\in\{0,1,\dots,9\}$ is the $n$-th decimal digit of the product in question. In addition, subtracting the above grid from $2a+2b+c-1$ yields the same grid reversed (with $m$ replaced by $c-1-m$), hence $r(2a+2b+c-1-n)=r(n)$ follows as well. This shows that the product in question is indeed palyndromic in base $10$.


