Thank you, Kerry I had in mind this little "pattern": 19, 29, 39, ... 89. Best, E. -------- Message d'origine-------- De: math-fun-bounces@mailman.xmission.com de la part de Kerry Mitchell Date: mer. 09/12/2009 17:44 À: math-fun Objet : Re: [math-fun] N/b gives 'a' as remainder 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) Best, É.
Here is one class of N: Let N = a + K*b, for integer K (the requirement that a is the remainder). For simplicity, let N also = a * 10^(2n) + b (concatenating a & b). Then, a * 10^(2n) + b = a + K*b, a(10^(2n) - 1) = b(K - 1) a(10^n + 1)(10^n - 1) = b(K - 1). Now, let K = 10^n, so both sides have a common factor. a(10^n + 1) = b. If the number of digits in a = n, then a & b work to give N. For example: - a = 1, b = 11, N = 111 - a = 25, b = 2525, N = 252525 - a = 2357, b = 23572357, N = 235723572357, etc. I'm sure someone much smarter than me can figure out a more general pattern. Kerry -- lkmitch@gmail.com www.kerrymitchellart.com http://spacefilling.blogspot.com/ _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun