[math-fun] N/b gives 'a' as remainder
Hello MathFun, If I split integer N into two substrings a and b, (for instance N=423156 --> a=423 and b=156) ... I have almost always b as a remainder of N/a. (This works also if I split N=423156 into a=42315 and b=6 ... or if I split N=423156 into a=4231 and b=56, etc.) The only exceptions for N are (I think): 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 23, 24, 25, 26, 27, 28, 29, 33, 34, 35, 36, 37, 38, 39, 44, 45, 46, 47, 48, 49, 55, 56, 57, 58, 59, 66, 67, 68, 69, 77, 78, 79, 88, 89, 99. (which are integers < 100 with equal digits or digits in increasing order) 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.
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/
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
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
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
participants (3)
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Andy Latto -
Eric Angelini -
Kerry Mitchell