Let d > 0 divide 99. I can then show that d divides n iff d divides the base-10 reverse of n. I believe the converse is true as well, that these are the only d with this property. In general, I believe that ( d | n <=> d | base-b reverse of n ) <=> d | b^2-1 Supposing this is indeed true, it can't be too hard to prove, but I'm at a loss. ----- No virus found in this message. Checked by AVG - www.avg.com Version: 10.0.1204 / Virus Database: 1435/3454 - Release Date: 02/19/11
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
So you have show that if d | b^2-1, => d | n iff d | n'. The murky part for me is the converse. For d not a divisor of b^2-1, my computer seems able to find n with d | n but not d | n'. For example, with b = 10, d = 2, we find d | 10 but not d | 01 = 1. Likewise, for b = 10, d = 999, we find d | 1009989 but not d | 9899001. I contend it is precisely the divisors d of b^2-1 with the desired property. How do we show this? ----- Original Message ----- From: "Adam P. Goucher" <apgoucher@gmx.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Sunday, February 20, 2011 8:05 AM Subject: Re: [math-fun] Digital question
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
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