Question

Suppose n is any positive integer having an even number of digits, and m is its reverse, for example if n = 1234, then m = 4321.
a) Show that m+n is always divisible by 11.
b) Why doesn't the same proof work for positive integers having an odd number of digits?

Answer 1

*bookmark* :D

Answer 2

This is based on idea of 10a+b+10b+a = 11(a+b) for 2 digit numbers..I guess u can proceed inductively.

You need pairs, so it won't work for odd (getting a counterexample for odd n is easy).

Answer 3

Okay, but, for 4 digit numbers, you have 1000a+b+1000b+a = 1001(a+b), and for 6 digits, you have 100001(a+b)... and so on. So how do you know that these numbers are divisible by 11 as well?

Answer 4

Not sure if I understand u, u mean how do I "know" 1001" is divisible by 11?

That's why I said proceed inductively, for 4 digits, 1001a +110b +110c +1001d, etc.

Looks like u want 10^(2n-1) +1 / 11 for all n, use induction.

Answer 5

I know the answer, I invented the question myself... I just want to see if you can write the induction proof.

Answer 6

Yes, I can write the induction proof, thats why I said @use induction@