I created A330562: Positive numbers k with property that if d is any nonzero digit of k then k mod d is also a digit of k. 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 101, 102, 103, 104, 105, 109, 110, ... and commented that k must always have a zero digit. Maybe someone could check? I did it in a hurry. On Sat, Dec 28, 2019 at 12:08 PM Éric Angelini <bk263401@skynet.be> wrote:
Hello Math-Fun, Numbers that, when divided by one of their digits, have another of their digits as remainder: S = 10,13,19,20,21,23,26,29,30,31,32,39,... Check: 10 = 10*1+0 13 = 4*3+1 19 = 2*9+1 20 = 10*2+0 21 = 10*2+1 23 = 7*3+2 26 = 4*6+2 29 = 3*9+2 30 = 10*3+0 31 = 10*3+1 32 = 10*3+2 39 = 4*9+3 ... It would be nice to see the numbers where this is true, whatever the dividing digit is chosen in them (except 0). We could call them NYE-numbers as 20191231 is such an integer. Check: 20191231 = 10095615*2+1 20191231 = 20191231*1+0 20191231 = 2243470*9+1 20191231 = 6730410*3+1
P.-S. There are also the integers where this routine is true for exactly two distinct digits, like 127 (not 133), as 127 = 63*2+1 and 127 = 18*7+1. Or exactly three distinct digits, four, etc. Best, É. (and forgive me, as usual, if this is old hat or if some typos are still present).
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