Even-length palindromic numbers mod 11 = 0

I was surprised to learn from a comment on Project Euler that palindromic numbers are always divisible by 11.

Here's the proof (for six digit numbers):
i.) A palindromic number has the form abccba [By definition]
ii.) abccba = 100000a + 10000b +  1000c + 100c + 10b + a [Expansion in Base 10]
iii.) abccba = (100000a + a) + (10000b + 10b) + (1000c + 100c) [Re-organizing ii.)]
iv.) abccba = 100001a + 10010b + 1100c [Re-organizing iii.)]
v.) abccba = 11 (9091a + 910b + 100c) [Factor out 11 from iv.)]
QED [= Proved, since we have factored 11 out.]


I think this always holds for palindromes (of any length) with an even number of characters, but it does not necessarily hold if you are allowed to include an 'odd' digit in the centre. (I know it sometimes holds because 121 = 11 x 11, but I don't know if 121 is a special case.)

Kaprekar's constant: 6174

I just discovered Kaprekar's constant, 6174, named after an Indian mathematician. As noted in the Wikipedia article on the constant, "This number is notable for the following property:

1. Take any four-digit number with at least two digits different. (Leading zeros are allowed.)
2. Arrange the digits in ascending and then in descending order to get two four-digit numbers, adding leading zeros if necessary.
3. Subtract the smaller number from the bigger number.
4. Go back to step 2.

The above operation, known as Kaprekar's operation, will always reach 6174 in at most 7 steps and it stops there. Once 6174 is reached, the process will keep yielding 7641 – 1467 = 6174."

Weird.

I am pleased to discover the Wikipedia has entries on many individual integers.

The image comes from Plus Magazine, a magazine of mathematics- click the image for a more detailed article on 6174.