4 ms·

Hm... 14 => 1 (3^1) + 4 (3^0) => 7 => divisible by 7. 21 => 6 (3^1) + 1 (3^0) => 7 => also divisible by 7. 7 * 53 = 371 => 3 (3^2) + 7 (3^1) + 1 (3^0) => 27

• card_zero 1y ago
  Hmm, no. There's a hazard of false positives here (which is why I had mistakes to fix previously). It's not all about the number 3; you don't want to multiply columns by 3², 3¹, 3⁰. Instead you want the remainder of the division of (ten to the column number) by seven (or whatever divisor you're checking). That's a repeating pattern. In the case of 3 the pattern has length 1, and it goes (1, 1, 1, ...). In the case of 7 the pattern has length 6, and it goes (1, 3, 2, 6, 4, 5), starting at the units column. If that looks a bit pseudorandom, there's a good reason, it's vaguely similar to an old-fashioned PRNG. (The pattern starts again with 1 in the millions column, because 142857 * 7 = 999999. Hence 2 million divided by 7 has remainder 2, 3 million divided by 7 has remainder 3, and so on.) So 371 should be understood as 3*2 + 7*3 + 1*1, which is 28. Or as another example, 6993 => 6*6 + 9*2 + 9*3 + 3*1, which is 84.
  • card_zero 1y ago
    Waaait a minute though. The results for (10 to the n) % 7 and the results for (3 to the n) % 7 are the same, so your version works precisely. That's because the difference is 7.