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$1, 2, 3, 4, 5, 6, 7, 8, X, 9, 10, 11, 12 ?$

What is the number x in the above series?

Hint: It's not 9.

Hint 2:

there is a reason that the series is a series of continuous from 1 to 8.

Hint 3:

there is something with the numbers of digits.

Hint 4:

position index. Here 8th term is 8 but 9th term can not be 9, instead the 10th term is 9.

Hint5:

after "x" the series again continues till it reaches the 99th term and then 999th term and so on..

Final Hint: No more hints after this (it was asked to us when i was in 7th grade)

98th and 99th terms are 97 each. 100th term is 98.

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    $\begingroup$ As show the answers, you should at least point where to think. Because there's indefinite number of approaches... $\endgroup$
    – nicael
    Jun 10, 2016 at 20:52
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    $\begingroup$ "I know, it's 10 because its the next number x for which 840/x is an integer!" $\endgroup$
    – palsch
    Jun 10, 2016 at 20:58
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    $\begingroup$ It's not 9, because 7-ate-9! So the next one must be 10 $\endgroup$
    – Inazuma
    Jun 11, 2016 at 3:21
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    $\begingroup$ number of natural numbers less than n+2 not in ( 10^m for whole m) $\endgroup$
    – Jasen
    Jun 12, 2016 at 2:39
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    $\begingroup$ the 9th term must be 8, 9+2=11, the numbers less than 11 not in 10^m are 2,3,4,5,6,7,8,9 - of which there are 8. $\endgroup$
    – Jasen
    Jun 13, 2016 at 12:43

5 Answers 5

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$\begingroup$

as this question hase ben taken off hold I'll post my answer

it's

8

because the sequence is

$$ s(n) = n-floor(log_{10}(n+1)) $$

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I think it is

nine, although you say it's not

because

Wolfram|Alpha says this and Wolfram|Alpha never lies. :P

Look:

Wolfram|Alpha response

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    $\begingroup$ Your infallible reference also believes that $\small\sf s,a,n,t,a,c,l,a,u,s \to 1/10 (3a{+}c{+}l{+}n{+}2s{+}t{+}u) = 1/6 (e{+}i{+}2s{+}t{+}x) \to e,x,i,s,t,s,...$ $\endgroup$
    – humn
    Jun 10, 2016 at 22:34
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    $\begingroup$ @humn Of course. W|A is always right, so he exists. Didn't you know? $\endgroup$
    – palsch
    Jun 11, 2016 at 11:28
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    $\begingroup$ @humn Santa Claus may exist! wolframalpha.com/input/?i=Does+Santa+Claus+exist%3F $\endgroup$
    – palsch
    Jun 11, 2016 at 14:20
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If it's not nine, than it's

ten, the numbers could be just the 9-base ;P

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  • $\begingroup$ I was just thinking this and then your answer popped up. :) +1 $\endgroup$
    – palsch
    Jun 10, 2016 at 20:38
  • $\begingroup$ @nicael there is a sequence which matches all and the next number.. can u justify this with 10? $\endgroup$
    – Wasiq Shahrukh
    Jun 10, 2016 at 20:41
  • $\begingroup$ What could else it be if you're just counting and there's no 9 in 9-base? $\endgroup$
    – nicael
    Jun 10, 2016 at 20:42
  • $\begingroup$ @nicael lateral thinking man $\endgroup$
    – Wasiq Shahrukh
    Jun 10, 2016 at 20:45
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    $\begingroup$ It's ten in base nine, what's wrong... $\endgroup$
    – nicael
    Jun 10, 2016 at 21:07
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What about

nothing ?

because

This are numbers of planets in the solar system, Mercury=1, Venus=2, Earth=3, ...

and

since 2006 there are only 8 left

so

It's not nine (anymore).

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Maybe it is

$0$? And the sequence are numbers modulo $9$.

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