On the GCD of two palindromes. I had an observation. Which I will discuss below. My question will be Is my observation correct? If so, how can one prove it?
Observation:
Consider the string of palindromes below:
$100...01$ and $111...11$
I observed that:
$gcd(100...01,111...11)=1$ if the length of the string is odd while
$gcd(100...01,111...11)=11$ if the length of the string is even.
Thank you so much for the big help.
 A: What I'm about to do below is called the Euclidean algorithm. If you're not familiar with it, I can show you some explanations of it on other questions/Web sites if you ask me in the comments.
We have:
$$\text{gcd}\left(10^k+1, \sum_{i=0}^k 10^k\right)$$
The second element is bigger, so subtract it by the first element:
$$\text{gcd}\left(10^k+1, \sum_{i=1}^{k-1} 10^k\right)$$
Notice how the second element now has no ones place or $10^k$ place. Now, the first element is bigger, so subtract it by the second element, which affects all of the digits of the first element except the first and the last:
$$\text{gcd}\left(91+\sum_{i=2}^{k-1} 8\cdot 10^k, \sum_{i=1}^{k-1} 10^k\right)$$
Notice how I changed the summation index from $i=1$ to $i=2$ and brought the tens place out into to make $91$ in front. Now, do this $8$ more times, getting rid of the summation and subtracting $91$ by $80$ which gives us $11$:
$$\text{gcd}\left(11, \sum_{i=1}^{k-1} 10^k\right)$$
Now, $11$ is prime so the GCD is either $1$ or $11$:


*

*If $k$ is even, then we have an odd number of $1$s in the second element and it is not divisible by $11$, so the GCD is $1$.

*If $k$ is odd, then we have an even number of $1$s in the second element and it is divisible by $11$, so the GCD is $11$.


Thus, since I just proved a restatement of your observation, your observation was indeed correct. Good job!
