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.

  • $\begingroup$ You certainly need some restrictions. $3t+1$ and $3t+2$ can't both be palindromes (unless you allow single digit numbers). Ah, I posted this before you edited in the gcd condition. $\endgroup$ – Gerry Myerson May 6 '15 at 12:35
  • $\begingroup$ Yes Sir @GerryMyerson, it must be that p and q are relatively prime. $\endgroup$ – Jr Antalan May 6 '15 at 12:36
  • $\begingroup$ Consider combining the systems into one. For n large and t small, pt + j is a palindrome implies (10^n + 1)(pt+j) is a palindrome. Consider how small t can be for both expressions to be palindromic, then pick n and m very large and with the right parity and consider pt+j ... qt+k ... qt+k ... pt+j. You can write this as a single recurrence that hopefully will produce a palindrome for some t depending on m and n. Gerhard "Give Yourself Plenty Of Room" Paseman, 2015.05.06 $\endgroup$ – Gerhard Paseman May 6 '15 at 18:28
  • $\begingroup$ Thanks @GerhardPaseman, sorry if I am having a hard time on understanding your comment, In particular what does $m$ represent and what is the meaning of combining the two system. As well as pt+j...qt+k...qt+k...pt+j. I am hopeful that with your help I can understand the answer to my query. Thanks again. $\endgroup$ – Jr Antalan May 6 '15 at 21:35
  • $\begingroup$ The idea is to choose m and n (and r) and consider the arithmetic progression (10^n+1)(pt+j) + (10^m +10^r)(qt+k). Although it won't be a palindrome for some choices of m and r and n, there is a lot of room to play in, and perhaps you can find ranges of t that guide your choice of m,n and r. If so, you have reduced two progressions to one, hopefully making it easier to study and predict. Gerhard "Maybe Consider Several Choices Simultaneously" Paseman, 2015.05.07 $\endgroup$ – Gerhard Paseman May 7 '15 at 23:14

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.

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  • $\begingroup$ Or ( converting string operations to arithmetic ) if t is such a number, so is (t - 27)/1000 . (Unless I need to remove the most significant digits instead.) $\endgroup$ – The Masked Avenger May 14 '15 at 23:29
  • $\begingroup$ Thanks Sir @TerryTao, I should have make the conditions for p and q more tight. That is I should have included that p and q must not be divisible by 10. Is it okay Sir to ask if the statement: There are infinitely many integers p and q and t satisfying the system for all j and k, being TRUE. Thanks again Sir. $\endgroup$ – Jr Antalan May 15 '15 at 2:00
  • $\begingroup$ One can still produce counterexamples. For instance, one cannot have $t$ and $2001t+1000$ both be palindromes, because then they would have the same first three digits, while having a ratio somewhere between 2001 and 3001, which one can easily check to be absurd. It might be a good exercise for you to experiment with further counterexamples of this type in case you wish to impose further restrictions on p,q,j,k. $\endgroup$ – Terry Tao May 15 '15 at 15:11
  • $\begingroup$ Yes Sir @TerryTao. Thanks a lot. Now I know the answer to my question. $\endgroup$ – Jr Antalan May 15 '15 at 21:51

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