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.
Best Answer
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.