The question is:

If 7 + 16 = 1000 what is 23 + 42?

Note, that I currently do not have the solution for this riddle, which is not publicly available.

I have already tried the following:

  • Input base 14 and output base 3, which would yield 10022

  • Base 987 calculation: 7 + 987*1 + 6 = 1000, 2*987+3+4*987+2 = 5927, which is also wrong.

Any further ideas? I will of course post the solution as soon as it is available!


closed as off-topic by ffao, Rand al'Thor, Wen1now, Deusovi Dec 8 '17 at 1:08

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  • 2
    $\begingroup$ I am marking as too broad. Especially since you don't have the answer, what is to say its not just left side + 977 or any other number of basic operations? $\endgroup$ – Jesse Dec 7 '17 at 23:44
  • $\begingroup$ @JesseBarnett Just assume it's not and the solution is more sophisticate, as it always is with good riddles. I presume, that it's good. $\endgroup$ – Ctx Dec 7 '17 at 23:46
  • $\begingroup$ with only one example we cannot establish a pattern, this just encourages people to come up with the most cool-looking solution when there is in fact an infinite number of solutions $\endgroup$ – Jesse Dec 7 '17 at 23:52
  • 3
    $\begingroup$ @JesseBarnett Maybe you should put some more effort in thinking about this riddle instead of finding excuses why you cannot solve it. $\endgroup$ – Ctx Dec 7 '17 at 23:53
  • $\begingroup$ Is unsolved-mysteries appropriate? Just because you don't have the solution doesn't mean nobody does... $\endgroup$ – Quintec Dec 8 '17 at 0:37

Not sure if right answer but if you think in binary you get

7 + 16 = 23 => 10111 Then the inverse = 1000

Do the same for 23 + 42

You get 23 + 42 = 65 => 1000001

And the Answer

And the inverse is 101110

  • $\begingroup$ I like this solution, but I fear, it is a bit too far fetched. I'll check if it works out, anyway. (+1 from me) $\endgroup$ – Ctx Dec 8 '17 at 1:04
  • 1
    $\begingroup$ Ok, this answer is not accepted for the riddle. I will accept your answer here anyway, since the question is on hold, hence there cannot be more answers, and I appreciate your effort. $\endgroup$ – Ctx Dec 8 '17 at 1:10

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