Common factors for all palindromes

Question

For example a palindrome of length $4$ is always divisible by $11$ because palindromes of length $4$ are in the form of: $$\overline{abba}$$ so it is equal to $$1001a+110b$$ and $1001$ and $110$ are divisible by 11

Is there a common factor for all palindromes of any length? if not how do you find the common factor of palindromes of a certain length?

Answer 1

As a counterexample for odd length, $121$ and $131$ are relatively prime. More generally,

$$1...121...1$$ and $$1...131...1$$ will always be relatively prime, since their difference will be of the form $2^k 3^j$ for some $k$ and $j$.

Answer 2

$11$ always divides palindromes of even length. There need not exist other common factors for other palindromes.

Answer 3

The greatest common divisor of all palindromes is $1$. It suffices to find two palindromes $a$ and $b$ with $\gcd(a,b)=1$.

Consider $a = 2$ and $b=3$. Clearly, $\gcd(a,b) = 1$.

A nontrivial counterexample: $\gcd(101,111)=1$.

Answer 4

$131$ is prime, so the common factor will be $1$ if odd length is allowed. All even length palindromes are divisible by $11$. You should be able to convince yourself that this is all.