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%I #30 Aug 31 2021 02:43:15
%S 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,
%T 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,
%U 6,9,11,11,1,4,3,1,1,1,1,1,1,1,1,1,1,9,7,7
%N Smallest number that when multiplied by n contains the digit 7.
%C Conjecture: 35 is the highest value to occur in this sequence.
%C 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
%H Robert Israel, <a href="/A285802/b285802.txt">Table of n, a(n) for n = 1..10000</a>
%e 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.
%p f:= proc(n) local k;
%p for k from 1 do
%p if has(convert(n*k,base,10),7) then return k fi
%p od
%p end proc:
%p map(f, [$1..100]); # _Robert Israel_, Nov 26 2019
%t Table[k = 1; While[DigitCount[k n, 10, 7] == 0, k++]; k, {n, 82}] (* _Michael De Vlieger_, Apr 26 2017 *)
%o (Python)
%o def a(n):
%o k=1
%o while True:
%o if "7" in str(n*k): return k
%o k+=1
%o print([a(n) for n in range(1, 101)]) # _Indranil Ghosh_, Apr 27 2017
%o (PARI) a(n) = {my(k=1); while(!vecsearch(vecsort(digits(n*k)), 7), k++); k;} \\ _Michel Marcus_, Jun 09 2018
%K nonn,base,easy,look
%O 1,1
%A _J. Lowell_, Apr 26 2017
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