5
$\begingroup$

Find a rule that applies for the first six entries and add at least one more.

I

do

take

20,000¥

recompensations 

The rule is simple.

Each individual entry has multiple valid solution, this is just one such set. The fact that the next entry is longer than the previous one doesn't need to be true for all entries.

$\endgroup$
  • 6
    $\begingroup$ Next words: in lieu of pre-determined donut payment. $\endgroup$ – Ian MacDonald Apr 20 '17 at 15:47
  • $\begingroup$ @IanMacDonald why would you think that? I'm curious. $\endgroup$ – Vepir Apr 20 '17 at 15:49
  • $\begingroup$ Is the trailing space after recompensations significant? $\endgroup$ – Scott M Apr 20 '17 at 15:49
  • $\begingroup$ @ScottM Yes it is. $\endgroup$ – Vepir Apr 20 '17 at 15:50
  • 9
    $\begingroup$ I cannot see any meaningful similarities amongst the words apart from them grammatically forming the beginning of a sentence when taken in the order provided. I have simply chosen to finish the sentence as though this were a court case involving a dispute about missing donuts. $\endgroup$ – Ian MacDonald Apr 20 '17 at 15:53
11
$\begingroup$

How about

at every opportunity that I can.

Each line

is double the byte count of the previous one.

$\endgroup$
  • $\begingroup$ Very Nice.Short and precise.Lovely. $\endgroup$ – Swarnabja Bhaumik Apr 20 '17 at 16:03
  • 5
    $\begingroup$ This leads me to believe that in lieu of agreed donut payment. would also be an acceptable solution. If that is the case, this puzzle is not specific enough and has far too many possible correct answers. $\endgroup$ – Ian MacDonald Apr 20 '17 at 17:10
  • 3
    $\begingroup$ @IanMacDonald The question is to find the rule. Which is unique. A hidden trailing space after the last entry hints toward the rule itself. It is stated that multiple valid solutions can replace the existing entries. $\endgroup$ – Vepir Apr 20 '17 at 18:22
  • 1
    $\begingroup$ @Vepir: But the rule isn't unique. $\endgroup$ – Deusovi Apr 20 '17 at 20:38
  • 1
    $\begingroup$ @Deusovi If you consider the fact we are dealing with bytes, which leads to binary, which leads to powers of 2, which is the rule; And the fact the the rule is simple. But I guess nothing is unique if you try to fit things like "a(n)=2a(n-1), except every tenth time you multiply by 1000/512 instead of by 2" ? or something like: a(1) = 1; for n>1, a(n) is the smallest positive integer not already present which is entailed by the rules (i) k present => 2k present; (ii) 3k+1 present and k odd => k present., // which is among "similar sequences". $\endgroup$ – Vepir Apr 20 '17 at 20:45

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