Reversed Digits Theorem

Date: 06/24/99 at 02:18:35
From: Anand  Anil Tikotekar
Subject: Number theory

Suppose abc... is a positive integer (where a, b, c, ... are its 
digits).

The theorem states:
     if                (abc...)^n = xyz...
     and if   (a + b + c + ...)^n = x + y + z + ...
     then              (...cba)^n = ...zyx

For example, consider the number 12. 12^2 = 144 and 21^2 = 441 (the 
opposite of 12 is 21 and the opposite of 144 is 441). This is because 
(1+2)^2 = 1+4+4. Another example is 13 (13^2 = 169 and 31^2 = 961, 
(1+3)^2 = 1+6+9.) In these examples n = 2. (Other examples are 12, 13, 
211, 212, 101, 11, 121, ...)

My question is, can you prove this theorem?

Date: 06/24/99 at 11:02:25
From: Doctor Rob
Subject: Re: Number theory

I'll show you how to do it with (ab)^2 = xyz, and you figure out how 
to do it in general.

We are assuming first of all that

     (10*a + b)^2 = 100*x + 10*y + z,

with a, b, x, y, and z all digits between 0 and 9, inclusive. That 
implies that z = b^2 - 10*p, y = 2*a*b + p - 10*q, and x = a^2 + q. 
p and q are the carries you get when you do the multiplication. Then

     x + y + z = a^2 + 2*a*b + b^2 - 9*(p + q),
               = (a + b)^2 - 9*(p + q).

Since we're assuming that x + y + z = (a + b)^2, it follows that 
p + q = 0. But p and q are carry digits, so they are nonnegative, 
and so this implies that p = q = 0.  Thus

     x = a^2   <= 9,
     y = 2*a*b <= 9,
     z = b^2   <= 9.

This implies that

   (10*b + a)^2 = 100*z + 10*y + x,

which is what you wanted to show.

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