Proof: Palindromes Divisible by 11

[Math100]

Homework Statement

A palindrome is a number that reads the same backward as forward (for instance, $373$ and $521125$ are palindromes). Prove that any palindrome with an even number of digits is divisible by $11$.

Relevant Equations

None.

Proof:

Suppose $N$ is a palindrome with an even number of digits.
Let $N=a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0}$, where $0\leq a_{k}\leq 9$, be the
decimal expansion of a positive integer $N$, and let $T=a_{0}-a_{1}+a_{2}-\dotsb +(-1)^{m}a_{m}$.
Note that $m$ is odd.
Then $N=a_{0}10^{m}+\dotsb +a_{m-1}10+a_{m}$.
This means $a_{i}=a_{m-i}$ for $0\leq i\leq m$.
Since $m$ is odd, it follows that $T=(a_{0}-a_{m})+(a_{2}-a_{m-1})+\dotsb +(a_{m-1}-a_{1})\implies 0+0+\dotsb +0=0$.
Thus $11\mid T\implies 11\mid N$.
Therefore, any palindrome with an even number of digits is divisible by $11$.

 

[fresh_42]

Looks ok, except that it is better to write at the end
... $T=(a_{0}-a_{m})+(a_{2}-a_{m-1})+\dotsb +(a_{m-1}-a_{1})= 0+0+\dotsb +0=0$.
with an equality sign instead of an implication sign.