8
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Am I thinking of a number?

Let us say I am, and the first number of which I am thinking is four.

The third number will be a number which is my favorite number.

And I'll hesitatingly call the second number The Culprit in this supposition: $$a=b \rightarrow \binom {(a+b)(a-b)} {(a-b)} = \binom {b}{(a-b)} \rightarrow 2a=b$$
Exciting, isn't it? That completes my first "set"



Now I move forward powerfully to my next "set," which is a single number that helps you color a map efficiently. That's the whole thing. One boring number!



Now I'll additionaly add another, third, final, ultimate, closing, concluding "set" that will terminate the end of all of the sets.

Digitally speaking, this last set is quite unimpressive.

It is, in fact, boring as hell to anyone who might be reading this, but it is very special to me.

Being consistent, it's again my favorite number, repeated six times. That is, it's a six-digit number with all six digits being the same.

As you surely can see, this results in something kind of magical, doesn't it?

Hint #1:

The sets have been determined by Powerful reasoning. Now, some Additional reasoning needs to take place.

Hint #2:

The tag says .

Hint #3:

So, we've found some numbers. And, using Mathematics, we might use those numbers in a number of ways. Perhaps these numbers together might form another number? Maybe you're feeling a bit numb from all of the numbers (bummer!), and, because you're a latecomer who realizes it's almost summer, you decide to slumber on a bed made from lumber, then POW, it all adds up to a new number!
You submit the answer, unencumbered, and treat yourself to a nice sandwich.
Perhaps cucumber?

Hint #4 (Hint #3 distilled):

Number from set 1
POW
Number from set 2
ADD
Number from set 3

Hint #5:

If you've followed the thread of this question all the way to this point, you may be asking yourself, "How is that ?"
And you'd be right in asking that, because, up to this (final) step, there has been none. (Hint, hint!)

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  • $\begingroup$ another hint plz;) $\endgroup$ – Preet May 2 '18 at 2:29
  • $\begingroup$ Read but don't get hint 3 - now feel dumber. *rimshot* $\endgroup$ – Rubio May 2 '18 at 23:59
  • $\begingroup$ @Rubio My apologies. Additional hint forthcoming. $\endgroup$ – Chowzen May 3 '18 at 0:51
  • $\begingroup$ Oh don't do that on my account, I was just having fun with the -umber theme. :) $\endgroup$ – Rubio May 3 '18 at 0:53
  • $\begingroup$ @Rubio 'sOK, it doesn't seem that anyone gets hint #3... (having trouble coming up with hints that don't give it away) $\endgroup$ – Chowzen May 3 '18 at 1:02
10
+50
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from all clues and Preet's answer I think the answer is

Mathemagics

How?

27171573562 is 11 digits long, same with the number of clues, choose the nth-letter from each clues where n is the digit from 27171573562 (2nd letter from first clue, 7th letter from second clue, etc.. ) will give you word mathemagics

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  • 1
    $\begingroup$ well done! I previously thought to pick the word instead of letter, then found nothing xD $\endgroup$ – athin May 4 '18 at 11:28
  • $\begingroup$ Wow that's a good riddle. Good job! $\endgroup$ – Yanko May 5 '18 at 9:39
7
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I believe the favorite number is:

6 as you described it with six different words(another, third, final, ultimate, closing, concluding) and its repeated six times as well. Or its six digit no. with all six digits being six;)

then final number could be(based on other answers):

4064666666 which looks like a mobile number. And the three sets are: 406|4|666666

Edit: three sets as think so but not sure:

three parts in number: country code, area code & prefix+line number. 406 code is for CHOTEAU city location

EDIT: Now what I get from hints:

(406 POW 4)+ 666666 = 27171573562

No idea what to do next

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  • 1
    $\begingroup$ This is very promising! So, why the three sets? 406 | 4 | 666666 $\endgroup$ – Chowzen Apr 20 '18 at 11:36
  • 1
    $\begingroup$ Please see Hint #1 $\endgroup$ – Chowzen Apr 23 '18 at 11:53
  • 2
    $\begingroup$ Huh! I just saw this post now (19 hours later). This newest 11 digit number is the next (and penultimate) step. $\endgroup$ – Chowzen May 3 '18 at 21:08
6
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I think that the second "set" is

4

And it refers to

The four colour theorem, which states that the minimum number of colours required to colour a map with no contiguous region painted with the same colour is 4.

No idea about the other "sets" though...

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5
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The second number (in the first set) is probably

0? As you divide by zero and get a "contradiction".

Edit (much later): Some notes on the "special" number

So we realized somehow that the riddle has something to do with the number $27171573562$. I realized two (possibly) important mathematical properties of this number. The first is the prime factorization $$27171573562=2 \times 7 \times 1940826683$$ The third prime is quit large. Another thing is, that this number is very similar (yet not exactly) to the numbers of the Euler constant $e=2.7182818284590452$

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  • $\begingroup$ The first conclusion is correct (+1). $\endgroup$ – Chowzen Apr 19 '18 at 16:28

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