The quotient of two palindromes

Question

Besides the multiples of 10, are there infinitely many positive integers which are not the quotient of two palindromes? Is 12 such an exception?

Answer 1

We wish to determine whether there exist positive palindromic integers $p$ and $q$ such that $12=\frac{p}{q}$, or equivalently, $12q=p$. That is, if we multiply $12$ by a palindrome can the product also be a palindrome?

Notation. The use of $x\dots x$ indicates any palindromic string of digits starting and ending with the digit $x$. For example: $1\dots1$ means any palindromic integer starting and ending with the digit $1$, including $1$ and $11$.

First, we note that when $12$ is multiplied by a palindrome of the form $1\dots1$ the last digit of the product will be $2$, when multiplied by $2\dots2$ the last digit will be $4$, and so on.

Then, to determine the possible first digits, consider the following sequence of inequalities:$$12=12\cdot1=12\cdot1=12\cdot1<12\cdot2=24$$ $$120=12\cdot10<12\cdot11=12\cdot11<12\cdot20=240$$ $$1200=12\cdot100<12\cdot101<12\cdot191<12\cdot200=2400$$ $$12000=12\cdot1000<12\cdot1001<12\cdot1991<12\cdot2000=24000$$ $$120000=12\cdot10000<12\cdot10001<12\cdot19991<12\cdot20000=240000$$ $$\vdots$$ These inequalities show that when we multiply $12$ by a palindrome of the form $1\dots1$ the first digit of the product can only be $1$ or $2$. We can use similar inequalities to determine the possible first digits when we multiply $12$ by $2\dots2$, $3\dots3$, etc.

The last column in the following table indicates whether the product can be a palindrome based on whether the first and last digits of the product can be the same when the given palindrome is multiplied by $12$. $$\begin{array}{|c|c|c|c|} \hline \text{Palindrome}&\text{First digit}&\text{Last digit}&\text{Possible?}\\ \hline 1\dots1&1\text{ or }2&2&\text{Yes}\\ \hline 2\dots2&2\text{ or }3&4&\text{No}\\ \hline 3\dots3&3\text{ or }4&6&\text{No}\\ \hline 4\dots4&4\text{ or }5&8&\text{No}\\ \hline 5\dots5&5\text{ or }6&0&\text{No}\\ \hline 6\dots6&6\text{ or }7&2&\text{No}\\ \hline 7\dots7&7\text{ or }8&4&\text{No}\\ \hline 8\dots8&8\text{ or }9&6&\text{No}\\ \hline 9\dots9&9\text{ or }1&8&\text{No}\\ \hline \end{array}$$ We need to analyse the palindrome $1\dots1$ further, so we now consider the first two digits and last two digits of the product, noting that the first two digits need to be the reverse of the last two digits to form a palindrome: $$\begin{array}{|c|c|c|c|} \hline \text{Palindrome}&\text{First digits}&\text{Last digits}&\text{Possible?}\\ \hline 10\dots01&12\text{ or }13&12&\text{No}\\ \hline 11\dots11&13\text{ or }14&32&\text{No}\\ \hline 12\dots21&14\text{ or }15&52&\text{No}\\ \hline 13\dots31&15\text{ or }16&72&\text{No}\\ \hline 14\dots41&16\text{ or }17&92&\text{No}\\ \hline 15\dots51&18\text{ or }19&12&\text{No}\\ \hline 16\dots61&19\text{ or }20&32&\text{No}\\ \hline 17\dots71&20\text{ or }21&52&\text{No}\\ \hline 18\dots81&21\text{ or }22&72&\text{No}\\ \hline 19\dots91&22\text{ or }23&92&\text{No}\\ \hline \end{array}$$ We conclude that $12$ can never be the quotient of two palindromes.

---

To consider whether there are infinitely many positive integers which are not the quotient of two palindromes we can apply the above approach to the infinite sequence of integers $1212,12012,120012,1200012$, etc.

For each of these integers, the argument proceeds as above. The tables need only minor changes and apply to all integers in the sequence. $$\begin{array}{|c|c|c|c|} \hline \text{Palindrome}&\text{First digit}&\text{Last digit}&\text{Possible?}\\ \hline 1\dots1&1\text{ or }2&2&\text{Yes}\\ \hline 2\dots2&2\text{ or }3&4&\text{No}\\ \hline 3\dots3&3\text{ or }4&6&\text{No}\\ \hline 4\dots4&4,5\text{ or }6&8&\text{No}\\ \hline 5\dots5&6\text{ or }7&0&\text{No}\\ \hline 6\dots6&7\text{ or }8&2&\text{No}\\ \hline 7\dots7&8\text{ or }9&4&\text{No}\\ \hline 8\dots8&9\text{ or }1&6&\text{No}\\ \hline 9\dots9&1&8&\text{No}\\ \hline \end{array}$$ Again, we need to analyse the palindrome $1\dots1$ further: $$\begin{array}{|c|c|c|c|} \hline \text{Palindrome}&\text{First digits}&\text{Last digits}&\text{Possible?}\\ \hline 10\dots01&12\text{ or }13&12&\text{No}\\ \hline 11\dots11&13\text{ or }14&32&\text{No}\\ \hline 12\dots21&14\text{ or }15&52&\text{No}\\ \hline 13\dots31&15\text{ or }16&72&\text{No}\\ \hline 14\dots41&16,17\text{ or }18&92&\text{No}\\ \hline 15\dots51&18\text{ or }19&12&\text{No}\\ \hline 16\dots61&19\text{ or }20&32&\text{No}\\ \hline 17\dots71&20\text{ or }21&52&\text{No}\\ \hline 18\dots81&21\text{ or }22&72&\text{No}\\ \hline 19\dots91&22,23\text{ or }24&92&\text{No}\\ \hline \end{array}$$

We conclude that there are infinitely many positive integers which are not the quotient of two palindromes.