2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $j$.

$\endgroup$
  • $\begingroup$ @user61527 I just have a comment about the numbers 1...121...1 and 1...131...1. How does it follow from the fact that their difference takes the form 2^k3^j for some k and j that they are relatively prime? Can you enligthen me on that matter? Thanks a lot. $\endgroup$ – Jr Antalan May 23 '16 at 9:42
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

$131$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.