1 = 1

2 = 12

3 = 27

4 = 44

5 = 35

6 = ?

The only hint I got was:

“Think outside the box”

  • 2
    $\begingroup$ Hi MA8, welcome to PSE! What is the source of the puzzle? $\endgroup$ – Greg Feb 5 at 18:39
  • $\begingroup$ Hi Greg, I actually have no reliable source for the puzzle. My friend gave it to me and he’s a puzzle addict, and I have been trying to figure the answer since three days. I gave up so I thought I could get help here! $\endgroup$ – MA8 Feb 5 at 18:50
  • $\begingroup$ Gotcha. No problem! Hope you get your answer! $\endgroup$ – Greg Feb 5 at 18:53
  • $\begingroup$ Could you perhaps list some of the possibilities or sequence rules you might have already tried? $\endgroup$ – visualnotsobasic Feb 5 at 19:15
  • $\begingroup$ I tried thinking of a mathematical formula for the sequence, none worked (I even used a computer to assist me). Then I came back to the hint and tried thinking about the words themselves: “one” “two” and so on, and I still can’t find a solution... But I am 90% sure it is not a mathematical sequence formula. $\endgroup$ – MA8 Feb 5 at 19:33

I think the answer might be

$6 = 816$


I think they are all Platonic numbers from a specific branch.

1 = 1, the first Platonic number in all branches
2 = 12, the second icosahedral number
3 = 27, the third cube number
4 = 44, the fourth octahedral number,
5 = 35, the fifth tetrahedral number.

Therefore, we must have
6 = 816, the sixth dodecahedral number
since this is the only remaining option.


"Think outside the box" could refer to the idea of thinking about Platonic numbers beyond cubes.

  • $\begingroup$ The values start with basically (1)Rank 4 -Dodecahedral and go up to (5) Rank 0 -Tetrahedral $\endgroup$ – GoldBishop Feb 6 at 15:03
  • $\begingroup$ So it could be either 146 or 816 with 6 starting the cycle over again or just reflecting back down the Ranks $\endgroup$ – GoldBishop Feb 6 at 15:07
  • $\begingroup$ This is the most answer that makes sense until now. Thank you very much! $\endgroup$ – MA8 Feb 6 at 19:11
  • $\begingroup$ Also...since all N forms have 1...it could be assumed that the sequence starts with Dodecahedral and goes up the r values...the only option after 5 is to either reflect or start all over from the "beginning $\endgroup$ – GoldBishop Feb 7 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.