A285802 Smallest number that when multiplied by n contains the digit 7.

7, 35, 9, 18, 14, 12, 1, 9, 3, 7, 7, 6, 6, 5, 5, 11, 1, 4, 3, 35, 7, 8, 9, 3, 3, 3, 1, 17, 3, 9, 7, 18, 9, 5, 2, 2, 1, 2, 2, 18, 7, 9, 4, 4, 6, 6, 1, 12, 3, 14, 7, 11, 7, 5, 5, 12, 1, 3, 3, 12, 7, 6, 6, 9, 11, 11, 1, 4, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 9, 7, 7

Offset

1,1

Comments

Conjecture: 35 is the highest value to occur in this sequence.

Conjecture is true. First of all we notice that a(10*k) = a(k), so we can consider only numbers not ending in 0. Then, it is easy to verify that for any number k between 1 and 99 not ending in 0, there is a multiple of k not larger than 35*k which has a 7 in one of its last two digits. Since the last two digits of the products only depend on the last two digits of k, this extends immediately to larger numbers. - Giovanni Resta, Apr 27 2017

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Example

a(2) = 35 because no even number ends in 7, but we can have even numbers whose next-to-last digit is 7; the smallest number is 70, which is even, and 2 times 35 is 70.

Maple

f:= proc(n) local k;
  for k from 1 do
    if has(convert(n*k, base, 10), 7) then return k fi
  od
end proc:
map(f, [$1..100]); # Robert Israel, Nov 26 2019

Mathematica

Table[k = 1; While[DigitCount[k n, 10, 7] == 0, k++]; k, {n, 82}] (* Michael De Vlieger, Apr 26 2017 *)

Prog

(Python)
def a(n):
    k=1
    while True:
        if "7" in str(n*k): return k
        k+=1
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Apr 27 2017
(PARI) a(n) = {my(k=1); while(!vecsearch(vecsort(digits(n*k)), 7), k++); k; } \\ Michel Marcus, Jun 09 2018

Crossrefs

Sequence in context: A196968 A000829 A247158 * A340523 A061825 A077536

Adjacent sequences: A285799 A285800 A285801 * A285803 A285804 A285805

Keyword

nonn,base,easy,look

Author

J. Lowell, Apr 26 2017

Status

approved