6
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One tough puzzle to solve :

Find the next number in the sequence :

5,105,74,712,37,?

Options are :

a. 2008
b. 57
c. 507
d. 98
e. 44

What is the next number, and why?

Puzzle Setter : Myself.

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closed as primarily opinion-based by JMP, Deusovi Aug 3 '16 at 1:56

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ I'm not sure why this question has received so many upvotes. As far as I can see, it's just another one of those questions where we have to spot something obscure which links the sequence. Like all of these questions, all of these answers can be justified one way or other - even by a polynomial. What's more, the 'number-theory' tag is misleading: "A mathematical puzzle whose solution is heavily based on the arithmetic properties of the integers." So our answers are going to be revolved around numbers since that's what we're expecting. $\endgroup$ – Shuri2060 Aug 2 '16 at 16:40
  • $\begingroup$ I don't see how slimeArmy's answer is any less justifiable than the one you had us looking for. $\endgroup$ – Shuri2060 Aug 2 '16 at 16:44
  • $\begingroup$ For example, I could say that the numbers in the sequence are the roots of this polynomial: $x^6 - 977x^5 + 213877x^4 - 18980531x^3 + 757923574x^2 - 12347718440x + 45032433600$ which would give me: e. 44 $\endgroup$ – Shuri2060 Aug 2 '16 at 16:48
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    $\begingroup$ QuestionAsker, Such a rule could be created for any sequence of numbers. I think that the point of the puzzle is to find a rule that is simpler (has a lower entropy) than the sequence and that lets us find the next item without using the options (it's a rule, not a criteria). $\endgroup$ – Runemoro Aug 2 '16 at 17:18
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    $\begingroup$ Had there been no options, There would be a huge number of answers to this question. $\endgroup$ – Sid Aug 2 '16 at 17:20
4
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I think the answer is

$D:98$

Because

If we write the numbers given out in English and count the characters used (including spaces) they all have a multiple of $4$ characters: $(4,20,12,24,12)$. Of the options only $D$ has a multiple of $4$ characters: $(22,11,22,\underline{12},10)$.

Another answer could be

$A:2008$

Because

If we write the numbers given out in English and remove spaces they are in alphabetical order, and only $2008$ would keep that order.

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  • $\begingroup$ Not sure how either fits the number-theory tag though :/ $\endgroup$ – Jonathan Allan Aug 2 '16 at 17:25
4
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The answer:

C: 507

Because:

None of the numbers in the sequence begin with an even number, which takes out option A and E.
The numbers also alternate between less than 100 and greater than 100, so the next number in the sequence would be greater than 100, removing B and D leaving C as the remaining option.

I'm looking for a more mathematical reason still, but is what I've conjured up as of now.

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0
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EDITED:

May be a long shot, but: c.507. If taking "0" as addition, like how we get 6 from 1+5, 5+7=12, which can be distributed on the two extra 7's to make 8 and 9, completing 0123456789. Cannot explain how I can distribute though.

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  • 1
    $\begingroup$ Sorry bro, not the right one. This is not an appropriate solution. Try it out. $\endgroup$ – Bharatwaaj Shankar Aug 2 '16 at 16:01
-3
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Ok, it's time for the answer. If you keep revolving around numbers then you won't be able to find the answer. It's lexicographically arranged numbers. Thus the answer is:

A: 2008

Thanks people for trying it out.

Five
One hundred and five
...
...
...
Thirty seven
Two thousand and eight.

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  • 5
    $\begingroup$ Man, you could have waited some more time. This puzzle has less than a hundred views and hence, the number of responses is less. $\endgroup$ – Sid Aug 2 '16 at 16:26
  • $\begingroup$ Haha. Let me try to give more puzzles. :D Thanks all. $\endgroup$ – Bharatwaaj Shankar Aug 2 '16 at 16:27
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    $\begingroup$ Shouldn't sequence puzzles have one and only one possible solution from N? $\endgroup$ – Erbureth Aug 2 '16 at 16:49
  • $\begingroup$ Welcome to Puzzling.SE. Please refrain from adding answers to your own question too soon, or even adding hints when not asked for. $\endgroup$ – ABcDexter Aug 2 '16 at 16:50
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    $\begingroup$ Doesn't seven hundred and twelve precede seventy-four lexicographically? $\endgroup$ – YowE3K Aug 3 '16 at 2:15

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