TITLE: A Question about Palindromic Numbers and System of Arithmetic Progression
QUESTION [8 upvotes]: Based from Harminc and Sotak's result, www.fq.math.ca/Scanned/36-3/harminc.pdf
We know that under certain condition, an arithmetic progression can contain an infinitely many palindromes.
My question will be, if I have a system of Arithmetic Progression such as
\begin{equation*}
3t+2\tag{1}
\end{equation*}
\begin{equation*}
4t+1\tag{2}
\end{equation*}
can I always find an integer $t$ such that $(1)$ and (2) are both palindromes? I know in my example that the answer is yes. But in general, if I have the system
\begin{equation*}
pt+j\tag{1}
\end{equation*}
\begin{equation*}
qt+k\tag{2}
\end{equation*}
with $\text{gcd}(p,q)=1$, can I find a $t$ for all $j<p$ and for all $k<q$ such that $(1)$ and $(2)$ are both palindrome? I am curious about this but I do not know how to answer it.
Or are there any reading materials that can help me on answering my query? Kindly help me.
Thanks for your help.

REPLY [11 votes]: Not in general.  For instance, there does not exist a natural number $t$ such that $t$ and $1000t+27$ (say) are both palindromes.  Indeed, if $1000t+27$ has the last three digits of $027$, hence has the first three digits of $720$ if it is a palindrome, hence $t$ has first three digits of $720$, hence $t$ has last three digits of $027$, hence $1000t+27$ has last six digits of $027027$.  Continuing this we see that $t$ consists entirely of strings of $720$ while also consisting entirely of strings of $027$, which is absurd.