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A354081 Positive integers k such that the first digit of k is divisible by the product of all the remaining digits of k. 0
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
LINKS
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
KEYWORD
nonn,base
AUTHOR
Moosa Nasir, Jun 05 2022
STATUS
approved

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Last modified March 28 05:39 EDT 2024. Contains 371235 sequences. (Running on oeis4.)