Many thanks to you too, Andy ! Best, E. -------- Message d'origine-------- De: math-fun-bounces@mailman.xmission.com de la part de Andy Latto Date: mer. 09/12/2009 20:15 À: math-fun Objet : Re: [math-fun] N/b gives 'a' as remainder On Wed, Dec 9, 2009 at 11:44 AM, Kerry Mitchell <lkmitch@gmail.com> wrote:
On Wed, Dec 9, 2009 at 9:12 AM, Eric Angelini <Eric.Angelini@kntv.be> wrote:
Now, are there integers N with N/b giving 'a' as remainder? (as usual, no integers a or b with leading zero -- like 09)
Assume b has k digits. Then we're looking for b and x such that a < b a * 10^k + b = xb + a a * (10^k -1) = (x-1) b. So all we have to do is choose b to be a k-digit proper factor of a(10^k-1) that is > a For example, with k = 1, a = 1, we want a factor of 9 > 1. So 13 and 19 work. k = 1, a = 4, we want 1-digit factors of 36 > 4. So 46, 49 both work. k = 2, a = 1, we want 2-digit factors of 99. So 133, and 199 work. k = 2, a = 4, we want 2-digit factors of 396. So 411, 422, 433, 444, 466, 499, 412, 436. Multidigit a's work, too. a = 24, k = 2, we want 2-digit factors of 24 * 99 > 24. So 2436. 2472, 2427, 2454, 2433, 2444, 2466, 2488. And so forth. Andy Latto andy.latto@pobox.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun