Reversed Digits TheoremDate: 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/ |
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