Problem #5: Palindromes

A number like 12321 is called a palindrome because it reads the same backwards as forwards. A friend of mine claims that all palindromes with four digits are exactly divisible by 11. Are they?

To start this problem, I began with smaller palindromes. Obviously the digits 1-9 are read the same forwards and backwards, but as you got into two digit (11, 22, 33..., 99) and three digit (111, 121, 131, ... , 999) they all seemed to be multiples of 11.

To look at 4 digit palindromes, I wanted to make a list of them (or at least see how many there are of them). I started my list and realized that there was a pattern:

Since there are only 90 of them, it wouldn't be completely absurd to try all of them, but I knew there had to be a more efficient way :)

I decided to just look at the first set (from 1001 to 1991) and see if I could find some sort of pattern. I actually tested all 10 of these and they were all divisible by 11. In writing these out I noticed that they were smaller numbers flipped around: 10 gave me 10 01, 11 gave me 11 11, and so on. Glancing up at my table, that would be the case for all of the numbers 10- 99 repeated and flipped. In writing out the first set I also noticed that the difference between all these numbers was 110 (a number also divisible by 11!)

As you can see this led me to creating an equation (the relationship had to be linear). For set number 1 it was 1001 + 110x where x was any number 1-9. Since 1001 and 110 are both divisible by 11, any multiple of 110 added to 1001 must be as well. I wrote out what set #2's equation would have been and found a way to relate it to set number one. Since 2002 is really 2*1001, I created a more generalized argument. Set #n could be figured out using 1001*n + 110*x where n and x are both any number between 1-9. Using my same logic that any multiple of a number divisible by 11 is allso divisible by 11 and the sum of any two numbers divisible by 11 is also divisble, we can conclude that this proof works.

As I was typing this I just thought of another way!! If we factor 1001*n + 110*x, we can pull out an 11 to get 11(91*n + 10*x) which obviously makes it divisible by 11!