- #1

Math100

- 771

- 219

- 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 ##.

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 ##.