14
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2, 3, 5, 7, ... 11, 13, 17, ... 25, 29, ... 35, ... 43, 47, 49, ... 53, 59, 61, 65, 67, ... 71, ... ??

  1. The answer (indicated by two question marks) has two digits.
  2. When you see '...', it means that one or more numbers are deliberately concealed at that point.
  3. The question is purely mathematical and some arithmetical calculation is needed.
  4. The numbers always increase as you move from left to right. The mathematical term for this is that the sequence is monotonically increasing.

Hint 0

A member submitted the following comment:
Under the assumption that 2 is the only even number in the sequence, we can say that the first "..." is 9, the fifth is 51, and the sixth is 69.
That statement is false and, in order to solve the puzzle as intended, you must understand why.


Hint 1

Ask yourself why there are gaps. What am I hiding?

Hint 2

It is an infinite sequence of integers, and 2 is the only even integer in the sequence.

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  • $\begingroup$ Under the assumption that 2 is the only even number in the sequence, we can say that the first "..." is 9, the fifth is 51, and the sixth is 69. $\endgroup$ – Ben Frankel Oct 17 '15 at 22:37
  • $\begingroup$ I guess that the dots are hiding non-integral numbers, and I guess that Hint 2 says that "2 is the only even integer in the sequence". Then the sequence should only contain one 2-digit integer number above 71 (that is, the ?? number). I have no idea how to find the sequence. $\endgroup$ – Gamow Oct 18 '15 at 15:06
  • $\begingroup$ @Gamow - It's an integer sequence. $\endgroup$ – chasly from UK Oct 19 '15 at 7:56
  • $\begingroup$ Are the integers represented in decimal? $\endgroup$ – Gamow Oct 19 '15 at 8:18
14
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The list contains:

the prime numbers in hexadecimal representation

The list then starts as:

2, 3, 5, 7, B, D, 11, 13, 17, 1D, 1F, 25, 29, 2B, 2F, 35, 3B, 3D, 43, 47, 49, 4F,
53, 59, 61, 65, 67, 6B, 6D, 71, 7F, 83, 89, 8B, 95, 97, 9D, A3, $\ldots,$ $\ldots,$ $\ldots$

Answer :

The answer to the puzzle is 83, the next term in the sequence given in the problem statement whose hexadecimal representation consists of two decimal digits.

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3
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The sequence is the union of the primes, the odd squares, and the multiples of 5 that are neither divisible by 3 nor 11, in increasing order.

So filling in the blanks and continuing the sequence,

2 3 5 7 .... 11 13 17 ........ 25 29 .. 35 ........ 43 47 49 .. 53 59 61 65 67 .. 71 .. ???

2 3 5 7 9 10 11 13 17 19 20 23 25 29 31 35 37 40 41 43 47 49 50 53 59 61 65 67 70 71 73 ...
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  • $\begingroup$ This fits every one of the requirements and it isn't too (?) arbitrary. Hopefully this isn't the intended answer. $\endgroup$ – Ben Frankel Oct 18 '15 at 5:33
  • $\begingroup$ You forgot the answer 'The answer (indicated by two question marks) has two digits.' $\endgroup$ – DrunkWolf Oct 18 '15 at 5:51
  • $\begingroup$ No. This isn't the intended answer. How do we proceed from here? Do I have to accept because you have given a specification -- or can I reject on account of its arbitrariness? How does one prevent 'union of sets' being the answer to every number sequence ever posed? Note: I have submitted a question on this very subject. When setting number-sequence puzzles, is there any accepted method to prevent arbitrary solutions? $\endgroup$ – chasly from UK Oct 18 '15 at 8:49
  • $\begingroup$ @chasly "When setting number-sequence puzzles, is there any accepted method to prevent arbitrary solutions" Yes, hint at clues in a backstory of sorts, this way the answer seems to fit and doesn't seem like it's 1 out of the million other answers that could fit without the backstory. $\endgroup$ – warspyking Oct 18 '15 at 12:37
  • 1
    $\begingroup$ @chaslyfromUK, You certainly can reject on account of arbitrariness. However, it is up to you to judge just how arbitrary it really is. If someone posts "this is the set {<exactly the numbers you included>, 99, 100, 101, ...}" then clearly the problem is the answerer and not the question. In this particular case however I would argue that the question should be improved somehow because at the moment my answer isn't entirely unreasonable. (suggestion: stop hiding the exceptional terms--those are exactly the numbers that would make solutions like this invalid). $\endgroup$ – Ben Frankel Oct 18 '15 at 20:53

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