Supposing this is indeed true, it can't be too hard to prove, but I'm at a loss.
If we concentrate on numbers with an odd number of digits, we have: n = a_0 + a_1 * b + a_2 * b^2 + a_3 * b^3 + ... + a_2k * b ^ 2k. The digital reverse is: n' = a_2k + a_2k-1 * b + ... + a_0 * b^2k And the difference is: n - n' = a_0 * (b^2k - 1) + a_1 * (b^2k-1 - b) + a_2 * (b^2k-2 - b^2) + ... Each of the numbers in parentheses is divisible by b^2 - 1. So, b^2-1 | n - n', hence n can be expressed as a linear combination of n' and b^2-1. From this, you can prove your conjecture (now theorem) for numbers with an odd number of digits. For an even number of digits, we can append an initial '0', and the same argument shows that d | b^2-1 ==> d | bn' and, as gcd(d,b) = 1, d | n'. Sincerely, Adam P. Goucher