Cody

# Jan Orwat

Rank
Badges
##### 23590
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51 – 100 of 3,602
 on 25 Nov 2016 on 21 Nov 2016 on 21 Nov 2016 on 21 Nov 2016 on 21 Nov 2016 on 21 Nov 2016 on 21 Nov 2016 Jan Orwat received Indexing II Master badge on 18 Nov 2016 on 18 Nov 2016 on 16 Nov 2016 on 16 Nov 2016 on 15 Nov 2016 on 15 Nov 2016 on 15 Nov 2016 on 15 Nov 2016 on 15 Nov 2016 on 15 Nov 2016 on 15 Nov 2016 on 15 Nov 2016 on 15 Nov 2016 on 15 Nov 2016 Jan Orwat submitted a Comment to Solution 1050679 You don't need integrator1, then change Gain on 11 Nov 2016 on 9 Nov 2016 on 9 Nov 2016 on 9 Nov 2016 on 9 Nov 2016 on 9 Nov 2016 on 9 Nov 2016 on 8 Nov 2016 on 8 Nov 2016 on 8 Nov 2016 Jan Orwat submitted a Comment to Solution 1047166 Thus it is not a 4-parasitic number. It doesn't meet the definition. Nevertheless is a great example of a cyclic number. You can get different cyclic permutations when multiply 142857 by 1, 2, 3, 4, 5 and 6. For 5 you got shift by one digit, therefore it's 5-parasitic. on 8 Nov 2016 Jan Orwat submitted a Comment to Solution 1047978 This solution shows a little bit different approach to the problem. It leads to nice, short code. on 8 Nov 2016 on 8 Nov 2016 on 8 Nov 2016 Jan Orwat received Indexing I Master badge on 8 Nov 2016 on 8 Nov 2016 Jan Orwat submitted a Comment to Solution 1047166 There is 4, not 5 on 8 Nov 2016 Jan Orwat submitted a Comment to Problem 1771. Polygonal numbers Explain why. It makes a lot of sense to me, and it's a lot easier (and makes even more educational sense) with 2016b. Just learn the difference between * and .* and ask yourself you need 1st or 2nd. on 15 Oct 2016 on 15 Oct 2016 on 10 Oct 2016 I have no idea what I was thinking of when I wrote this. Thanks for pointing that out so quickly. on 8 Oct 2016 on 7 Oct 2016 on 7 Oct 2016 on 6 Oct 2016 on 6 Oct 2016 on 6 Oct 2016
51 – 100 of 3,602