142857

  #!/usr/bin/env python
  
  import random
  
  digits = [1, 4, 2, 8, 5, 7]
  
  for times in range(20):
      digits_left = digits[:]
      nums = []
      while digits_left:
         this_num = 0
         for i in range(random.randint(1, len(digits_left))):
             digit = random.choice(digits_left)
             this_num *= 10
             this_num += digit
             digits_left.remove(digit)
         nums.append(this_num)
      print " + ".join(map(str, nums)), "=", sum(nums), "(a multiple of 9)"

  4271 + 58 = 4329 (a multiple of 9)
  7412 + 58 = 7470 (a multiple of 9)
  8 + 124 + 5 + 7 = 144 (a multiple of 9)
  845127 = 845127 (a multiple of 9)
  147285 = 147285 (a multiple of 9)
  1 + 824 + 75 = 900 (a multiple of 9)
  5 + 2 + 481 + 7 = 495 (a multiple of 9)
  758 + 41 + 2 = 801 (a multiple of 9)
  47125 + 8 = 47133 (a multiple of 9)
  7584 + 12 = 7596 (a multiple of 9)
  1275 + 8 + 4 = 1287 (a multiple of 9)
  475 + 12 + 8 = 495 (a multiple of 9)
  12 + 5 + 478 = 495 (a multiple of 9)
  4251 + 7 + 8 = 4266 (a multiple of 9)
  185 + 274 = 459 (a multiple of 9)
  28514 + 7 = 28521 (a multiple of 9)
  41278 + 5 = 41283 (a multiple of 9)
  457182 = 457182 (a multiple of 9)
  845712 = 845712 (a multiple of 9)
  2718 + 54 = 2772 (a multiple of 9)

So that, then, is the fascinating root observation: that any series of decimal number made from these digits is a multiple of nine.

What this means is that we can choose six random powers of 10 between 100 and 105 and make vector out of them, for instance <1, 100, 10, 10, 100000, 1000>. Then we do a dot-product between this and the vector <1, 4, 2, 8, 5, 7>. The result will be a multiple of 9.

If we choose the vector as <1, 1, 1, 1, 1, 1> we get the straight sum of the digits. If we choose <100000, 10000, 1000, 100, 10, 1> we get 142857, and so on.

Here is why it works:

   10**<whatever> x == x (mod 9).

For instance 4 mod 9 == 4. 40 mod 9 == 4. 400 mod 9 == 4.

And the reason for that is that 10 is 9 + 1; i.e. 10 is congruent to 1 modulo 9. So we are really multiplying by 1 under the congruence.

So the choices of powers of ten in the coefficient vector do not matter.

It works in other bases. For instance if any integer x is multiplied by a power of 6, that is congruent to x, modulo 5:

  $ clisp -q
  [1]> (mod (* 2) 5)
  2
  [2]> (mod (* 6 2) 5)
  2
  [3]> (mod (* 6 6 2) 5)
  2
  [4]> (mod (* 6 6 6 2) 5)
  2

Okay, I now wrapped my hackerly head around this enough that I can garbage collect it away as fairly uninteresting.

:)

But, one more thing: what is special about 1, 4, 2, 8, 5, 7 in connection to 9? Why, they are relatively prime to 9. The remaining three positive residues in the mod 9 congruence are not: 0, 3 and 6.

See here: https://en.wikipedia.org/wiki/Euler%27s_totient_function

Euler's totient function phi counts the number of such integers. phi(9) == 6 (there are six of these numbers for 9). The above page even uses this very example.

1, 4, 2, 8, 5, 7 comprise the multiplicative group of integers modulo 9.

https://en.wikipedia.org/wiki/Multiplicative_group_of_intege...

And yeah, that's pretty much the point of eliding bits of text, to put things together that weren't together before; I'm almost positive that's why we invented that specific punctuation mark.

It's like saying: "The 35th president of the United States, JFK, who was widely known for the space race and the bay of pigs, was elected after the 34 presidents before him."

And my comment is like: "The 35th president of the United States ... was elected after the 34 presidents before him."