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You are given the following 4 by 4 grid of numbers. Create four groups of four numbers such that the numbers in each group share some common property.

number connections no. 2. (648 388 407 577 777 120 333 216 384 592 466 144 125 693 246 296)

This puzzle is a (hopefully harder) sequel to the original number connections puzzle.

Hint 0 (Update):

As Nis Jørgensen correctly identified, the first group consists of 144, 216, 384, and 648 - being numbers of the form $2^a3^b$.

Hint 1:

In fact, I could replace 125 with a certain number $x$, and the same solution I had in mind (swapping out 125 for $x$, of course), would work just fine.

Hint 2:

Oh, and did I mention that the value of $|125-x|$ in Hint 1 is quite small?

Hint 3:

It's not all about number theory! Mathematics is more than just number theory! It can be much more real or much more complex than that! It can exponentially ... er ... bifurcate into more advanced fields! In the end it's up to you to make the final call based on your own judgment.

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  • $\begingroup$ I see gcd(407, 777, 333, 592, 296)=37, but it doesn’t satisfy the four number condition. $\endgroup$ Commented Jul 12 at 0:24
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    $\begingroup$ @RDK that is intended; one of them is the odd one out :) $\endgroup$ Commented Jul 12 at 1:18

3 Answers 3

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Group Numbers Connection
1 144, 216, 384, 648 3-smooth numbers (numbers of the form ($2^a3^b$)) — matches Hint 0
2 333, 407, 592, 777 37-Harshad numbers
3 120, 466, 577, 693 Appear as first three digits of notable mathematical constants — matches Hint 3
4 125, 246, 296, 388 Ferrari model numbering conventions (consistent with Hint 1 and the “small change to 125” clue)
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  • $\begingroup$ Not quite. As a hint, the first two of your categories (Harshad and div 24) are both one away. $\endgroup$ Commented Jul 10 at 18:22
  • $\begingroup$ well at least this is valid answer :D $\endgroup$ Commented Jul 10 at 18:37
  • $\begingroup$ yes, albeit different from the one I had in mind, curious to see if anyone can find it :) $\endgroup$ Commented Jul 10 at 18:54
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    $\begingroup$ Your answer fails the criteria that the categories must determine the answer. 144 and 384 can be swapped in your solution. Also, I think that using a category that has half of all integers[] in it makes for a pretty uninteresting puzzle. My hope would be that the answer is cleverer. [] Using an intuitive definition of "half". $\endgroup$ Commented Jul 11 at 11:49
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Here are some possible partial answers:

144, 216, 384 and 648 have 2 and 3 as their only prime factors

246,388,466, 577 each contain a unique prime factor - one not shared with any other number in the grid (41,97,233 and 577, respecitely).

296, 333, 407, 592 and 777 are all divisible by 37. This seems notable. If we can find a connection between the last 3, which also applies to one of these, we have a solution. As noted by Oray 333, 407, 592 and 777 are Harshad numbers. The OP has indicated that this is "one away", and thus 3 of these are in a category together.

Finally

120, 125 and 693 are the numbers left over. I have nothing.

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  • $\begingroup$ I believe you mean 777 instead of 577 being divisible by 37. the first category (144, 216, 384, 648) is correct. $\endgroup$ Commented Jul 11 at 22:42
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Here's a full solution which works, with no swapping possible, and simple-to-understand categories (no 3rd degree Wilson numbers divisible by 17 or the like).

120, 125, 144, 216 - integers in the interval 1-216

246, 296, 333, 384 - integers in the interval 246-384

388, 407, 466, 577 - integers in the interval 388-577

592,648 693,777 - integers in the interval 592-777

Finally:

I'll show myself out ...

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  • $\begingroup$ At least this is valid answer :D $\endgroup$ Commented Jul 16 at 10:43

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