5
$\begingroup$

1 = 1

2 = 12

3 = 27

4 = 44

5 = 35

6 = ?

The only hint I got was:

“Think outside the box”

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10
  • 2
    $\begingroup$ Hi MA8, welcome to PSE! What is the source of the puzzle? $\endgroup$ Feb 5, 2019 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, 2019 at 18:50
  • $\begingroup$ Gotcha. No problem! Hope you get your answer! $\endgroup$ Feb 5, 2019 at 18:53
  • $\begingroup$ Could you perhaps list some of the possibilities or sequence rules you might have already tried? $\endgroup$ Feb 5, 2019 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, 2019 at 19:33

1 Answer 1

3
$\begingroup$

I think the answer might be

$6 = 816$

Reasoning

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.

Hint

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

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4
  • $\begingroup$ The values start with basically (1)Rank 4 -Dodecahedral and go up to (5) Rank 0 -Tetrahedral $\endgroup$
    – GoldBishop
    Feb 6, 2019 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, 2019 at 15:07
  • $\begingroup$ This is the most answer that makes sense until now. Thank you very much! $\endgroup$
    – MA8
    Feb 6, 2019 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, 2019 at 18:44

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