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 A348833 Largest remainder of an n-digit zeroless number when divided by its product of digits. 2
 0, 42, 483, 4894, 47510, 468558, 3286509, 33038431, 311252462, 3026431197, 24615266110, 246854569382, 2374402515012, 18660293493470, 176435471621403, 1352572452024739, 16614091501667154, 99621458812003515, 810999063879719306, 7843786633582522125, 76818560735237400564 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Inspired by A348730 where product is replaced by sum. The largest product of digits that an n-digit number can reach is 9^n, according to this, a(n) < 9^n; indeed a(n) < 10^n/2. Corresponding n-digit numbers which have the largest possible remainder when divided by their digit product are in A348834. LINKS Chai Wah Wu, Table of n, a(n) for n = 1..53 EXAMPLE 5 == 0 mod (5). 96 == 42 mod (9*6). 969 == 483 mod (9*6*9). 9997 == 4894 mod (9*9*9*7). 99998 == 47510 mod (9*9*9*9*8). MATHEMATICA a[n_] := Max[Mod[#, (Times @@ DeleteCases[IntegerDigits[#], 0])] & /@ Range[10^(n - 1), 10^n - 1]]; Array[a, 7] (* Amiram Eldar, Nov 01 2021 *) PROG (Python) from math import prod from itertools import product def a(n): maxr = 0 for p in product("123456789", repeat=n): maxr = max(maxr, int("".join(p))%prod(map(int, p))) return maxr print([a(n) for n in range(1, 7)]) # Michael S. Branicky, Nov 01 2021 CROSSREFS Cf. A002473, A004526, A052382, A007602, A348730, A348834. Sequence in context: A216111 A216113 A204566 * A249233 A249234 A250324 Adjacent sequences: A348830 A348831 A348832 * A348834 A348835 A348836 KEYWORD nonn,base AUTHOR Bernard Schott, Nov 01 2021 EXTENSIONS a(9) from Giorgos Kalogeropoulos, Nov 01 2021 a(10)-a(12) from Michael S. Branicky, Nov 01 2021 a(13)-a(21) from Chai Wah Wu, Nov 08 2021 STATUS approved

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Last modified May 29 05:33 EDT 2024. Contains 372921 sequences. (Running on oeis4.)