A354081 Positive integers k such that the first digit of k is divisible by the product of all the remaining digits of k.

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 21, 22, 31, 33, 41, 42, 44, 51, 55, 61, 62, 63, 66, 71, 77, 81, 82, 84, 88, 91, 93, 99, 111, 211, 212, 221, 311, 313, 331, 411, 412, 414, 421, 422, 441, 511, 515, 551, 611, 612, 613, 616, 621, 623, 631, 632, 661, 711, 717, 771

Offset

1,2

Links

Table of n, a(n) for n=1..60.

Example

9331 is a term: the first digit is 9, which is divisible by the product of the remaining digits, i.e., 3*3*1 = 9.
8448 is not a term: the first digit is 8, which is not divisible by the product of the remaining digits, i.e., 4*4*8 = 128.

Mathematica

Select[Range[1000], !MemberQ[d = IntegerDigits[#], 0] && Divisible[First[d], Times @@ Rest[d]] &] (* Amiram Eldar, Jun 09 2022 *)

Prog

(C#)
public static bool a(int n)
{
   int product = 1;
   while (n > 9)
   {
      product *= n % 10;
      n /= 10;
      if (product == 0 || product > 9) { return false; }
   }
   if (n % product == 0) { return true; } else { return false; }
}
(PARI) isok(k) = my(d=digits(k), p=vecprod(d)); p && ((d[1] % (p/d[1])) == 0); \\ Michel Marcus, Jun 06 2022
(Python)
from math import prod
def ok(n):
    d = list(map(int, str(n)))
    return 0 not in d and int(d[0])%prod(d[1:]) == 0
print([k for k in range(800) if ok(k)]) # Michael S. Branicky, Jun 09 2022

Crossrefs

Subsequence of A052382.

Sequence in context: A038724 A106001 A161390 * A096106 A321990 A076641

Adjacent sequences: A354078 A354079 A354080 * A354082 A354083 A354084

Keyword

nonn,base

Author

Moosa Nasir, Jun 05 2022

Status

approved