A348833 Largest remainder of an n-digit zeroless number when divided by its product of digits.

0, 42, 483, 4894, 47510, 468558, 3286509, 33038431, 311252462, 3026431197, 24615266110, 246854569382, 2374402515012, 18660293493470, 176435471621403, 1352572452024739, 16614091501667154, 99621458812003515, 810999063879719306, 7843786633582522125, 76818560735237400564

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