Solution e68c02c5-af27-46a8-882e-eb1ce7cbf2e6

Points: 2 pts

Lemma 1: $11\mid P_2$

Let a palindrome with $2n$ digits be denoted by $P_{2n}$. Then,
\[
P_2=\overline{AA}
\]
for some integer $A\in\{1,2,\ldots,9\}$. Clearly,
\[
11\mid \overline{AA}=A\cdot 11.
\]
Hence, the base case of our induction is satisfied.

Lemma 2: $11\mid P_{2(n+1)}$ given $11\mid P_{2n}$

Given $P_{2n}$ with the form
\[
P_{2n}=\overline{a_n a_{n-1}\ldots a_{n-1}a_n},
\]
we can form $P_{2(n+1)}$ using
\[
P_{2(n+1)}=10P_{2n}+\left(A\cdot10^{2n+1}+A\right),
\]
because the highest power of $10$ in $P_{2n}$ is $10^{2n-1}$. By our inductive hypothesis,
\[
P_{2n}\equiv 0\pmod{11},
\]
and we see
\[
P_{2(n+1)}\equiv 0+A(-1)^{2n+1}+A\equiv 0\pmod{11}.
\]
Hence,
\[
11\mid P_{2(n+1)}.
\]

Final answer:
\text{TRUE}