Four-digit Palindromes Divisible by 11

Date: 02/10/99 at 20:24:01
From: Bonnie McCommons
Subject: Palindromes

I am asking in response to a question posed in a elementary education 
math class for which a definite answer was not arrived at by the 
class. Can you help us out?

Why are four-digit palindromes divisible by 11?

Date: 05/26/99 at 13:31:06
From: Doctor Bonnie
Subject: Re: Palindromes

This is a very interesting question and it took a while to figure it 
out. First let me remind you of some stuff to keep in mind as you read 
the proof:

Don't forget what it means to 'borrow' when you subtract. Suppose we 
subtract 93 - 18. W have to borrow from the nine and make the three a 
13 in order to be able to subtract 8, right? The 13 is actually 10 + 
3.

Since I have to write this in an e-mail, I will do all processes of 
long division according to DMSB: Divide, Multiply, Subtract, Bring 
down. Also note that stuff written like (B-A)(A) is a two-digit 
number where the first digit is (B-A) and the second digit is A. If 
you write out the process in long division, it will be easy to see 
what is going on.

Okay, let's suppose we have a four-digit palindrome. We can represent 
the digits using A and B: ABBA. Note that both A and B are less than 
or equal to nine and greater than or equal to zero. Now, either A<B or 
B<A or they are equal. If A is less than or equal to B then the 
following holds:

Dividing 11 into (A)(B)(B)(A):
D: 11 into (A)(B) goes A times 
M: 11 times A is (A)(A)
S: (A)(B) minus (A)(A) is (B-A)
B: bring down B

D: 11 goes into (B-A)(B), B-A times (because B-A < B)
M: 11 times B-A is (B-A)(B-A)
S: (B-A)(B) minus  (B-A)(B-A) is B-(B-A) which is A
B: bring down A

D: 11 goes into (A)(A), A times
M: 11 times A is (A)(A)
S: (A)(A)-(A)(A) is zero and we are done.

Now suppose A is less than B. We have to break this further into 
cases. First suppose B = A-1; then the following holds:

Dividing 11 into (A)(B)(B)(A):
D: 11 into (A)(B) goes A-1 times (the way 11 would go into 83 seven 
   times, i.e. 8-1 times)
M: 11 times A-1 is (A-1)(A-1)
S: (A)(B) minus (A-1)(A-1) is (1)(0) (ten)
B: bring down B

D: 11 goes into (1)(0)(B), 9 times
M: 11 times 9 is (9)(9)
S: (1)(0)(B) minus  (9)(9) is (10+B-9) which is B+1, which is A (this 
   is where we borrowed from the (0) and then the (1), which is how we 
   get 10+B)
B: bring down A

D: 11 goes into (A)(A), A times
M: 11 times A is (A)(A)
S: (A)(A)-(A)(A) is zero and we are done.

Now suppose B < A-1; then the following holds:

Dividing 11 into (A)(B)(B)(A):
D: 11 into (A)(B) goes A-1 times 
M: 11 times A-1 is (A-1)(A-1)
S: (A)(B) minus (A-1)(A-1) is (10+B-[A-1])  (again, we borrowed from 
   the A making it A-1 to give us 10+B so we could subtract)
B: bring down B

D: 11 goes into (10+B-A+1)(B), 10+B-A+1-1 times (note here that since 
   10+B-A+1 = B+11-A, this is still a single digit. Also it is less 
   than B, since 11-A>1 [remember that A is a digit and so it cannot 
   be bigger than 9])
M: 11 times 10+B-A+1-1 is (10+B-A+1-1)(10+B-A+1-1)
S: (10+B-A+1)(B) minus (10+B-A+1-1)(10+B-A+1-1) is (10+B-[10+B-A+1-1]) 
   which is A (once again do the borrowing thing)
B: bring down A

D: 11 goes into (A)(A), A times
M: 11 times A is (A)(A)
S: (A)(A)-(A)(A) is zero and we are done.

Okay? This is a bit detailed and may look somewhat confusing, but if 
you go through the long division calmly, it should make itself clear. 
Also, if you get hung up in the As and Bs, use numbers a while; that 
should get you back on track. If you need any further clarification, 
please write back. Also, don't forget to write in with any other 
questions. Good luck!

- Doctor Bonnie, The Math Forum
  http://mathforum.org/dr.math/