Proof: Palindromes Divisible by 11

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Any palindrome with an even number of digits can be expressed in a specific decimal format, leading to a calculation of T, which represents the alternating sum of its digits. Since the digits of a palindrome are symmetric, T simplifies to zero, demonstrating that T is divisible by 11. Consequently, this implies that the palindrome itself is also divisible by 11. The proof highlights the importance of the structure of palindromes in establishing divisibility. Thus, it concludes that all even-digit palindromes are divisible by 11.
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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 ##.
 
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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.
 
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