7 ms·

Hey neat, my favorite number is on HN! Here's an interesting video showing this stuff (And more) in a more user-friendly way https://www.youtube.com/watch?v=WU

• rbonvall 11y ago
  It's also my favorite number, but I don't think any of us likes it more than the guy in the video :) Thanks for sharing. I learned about 142857 when reading The Man Who Counted as a kid. I remember getting obsessed with it and discovering more properties (much fewer than the ones shown on the video), e.g.: 142 + 857 = 999 14 + 28 + 57 = 99
  • zyrthofar 11y ago
    You can go even further. 142 + 857 = 999; 9 + 9 + 9 = 27; 2 + 7 = 9 14 + 28 + 57 = 99; 9 + 9 = 18; 1 + 8 = 9 1 + 4 + 2 + 8 + 5 + 7 = 27; 2 + 7 = 9 Also, 1428 + 57 = 1485; 1 + 4 + 8 + 5 = 18; 1 + 8 = 9 14 + 2857 = 2871; 2 + 8 + 7 + 1 = 18; 1 + 8 = 9 1 + 42857 = 42858; 4 + 2 + 8 + 5 + 8 = 27; 2 + 7 = 9 1 + 42857 = 42858; 42 + 858 = 900; 9 + 0 + 0 = 9 This is kinda creepy. I wasn't really expecting the "also" additions to work.
    • schoen 11y ago
      If you have any multiple of 9 and repeatedly add the digits, you get back to 9: https://en.wikipedia.org/wiki/Digital_root https://en.wikipedia.org/wiki/Digital_root Conversely, if you have any number whose digits added together sum to 9 (or a multiple of 9), the original number is a multiple of 9. So, all of your original sum numbers (999, 99, 27, 1485, 2871, and 42858) are themselves multiples of 9, which will be the case for any number obtained by adding a set of numbers which together contain all and only the digits 142857. That means a lot of other "also" additions will work out too! You can even change the order, like 578 + 214 = 792 (a multiple of 9, and hence the digital root will end up being 9). Any order and any choice of how to break the numbers will work, because of the digital root property (and, importantly, the rule that "The digital root of a + b is congruent with the sum of the digital root of a and the digital root of b modulo 9").
      • schoen 11y ago
        Try this fun Python program to see how multiples of 9 are always generated no matter how you combine the digits: #!/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)" Sample output: 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)
      • acheron 11y ago
        I figured we all learned this from Square One: https://www.youtube.com/watch?v=Q53GmMCqmAM https://www.youtube.com/watch?v=Q53GmMCqmAM
      • kazinator 11y ago
        > multiples of 9, which will be the case for any number obtained by adding a set of numbers which together contain all and only the digits 142857 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). That is to say, any integer x is congruent, modulo 9, to a power of 10 times that integer. 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 Elementary number theory, my dear Watson. 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 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_integers_modulo_n https://en.wikipedia.org/wiki/Multiplicative_group_of_intege...
    • blaincate 11y ago
      you forgot 14285 + 7 = 14292; 1+4+2+9+2 = 18; 1+8 = 9