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I recently had a IQ Test taken and we all got stuck on the same question. The question was:

What comes next in the following sequence? $$58, 26, 16, 14,\_\_$$

The answer given in the answer sheet was $10$. My question is why? What pattern exists in those numbers?

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5
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    $\begingroup$ see also: bit.ly/SZZSxA $\endgroup$
    – wim
    Oct 17, 2012 at 11:23
  • $\begingroup$ I did google it, and went through 16 pages of results and couldn't find anything. $\endgroup$ Oct 17, 2012 at 20:44
  • $\begingroup$ it's the first result!!! $\endgroup$
    – wim
    Oct 17, 2012 at 23:53
  • $\begingroup$ @wim Ironically, this question is now the first result for that search query ;) $\endgroup$ Oct 12, 2013 at 3:29
  • $\begingroup$ math.stackexchange.com/questions/1814676/… $\endgroup$
    – pHotone
    Jun 6, 2016 at 5:17

6 Answers 6

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$5+8 = 13$ and twice $13$ is $26$, etc.

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  • $\begingroup$ i am still confused?!? can you please explain in more details thanks. .... $\endgroup$
    – user44961
    Oct 17, 2012 at 6:18
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    $\begingroup$ $5+8=13$ and twice 13 is 26. $2+6=8$ and twice 8 is 16. $1+6=7$ and twice 7 is 14. $1+4=5$ and twice 5 is 10. $\endgroup$
    – hpesoj626
    Oct 17, 2012 at 6:24
  • $\begingroup$ Amazing sequence... ! $\endgroup$
    – mCasamento
    Oct 17, 2012 at 8:40
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    $\begingroup$ not sure what this has to do with intelligence, though! $\endgroup$
    – wim
    Oct 17, 2012 at 11:19
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    $\begingroup$ If, in Gardner's theory, this question is supposed to be relevant to Maths/Logic intelligence, then Gardner and I have very different ideas of what Maths/Logic means. $\endgroup$ Oct 11, 2013 at 15:13
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I will probably be accused of pedantry for this answer, but I cannot help but point out that a question like this is not a math question. There are many sequences that begin 58, 26, 16, 14, 10. There are many patterns in the sequence 58, 26, 16, 14, 10, by which I mean that there are many computer programs that compute a sequence that begins this way, and they do not all finish the sequence in the same way.

I understand that what the OP is looking for is the simplest pattern in the sequence, which gives the simplest continuation of the sequence. My objection is that "simple" here has no mathematical meaning (Kolmogorov complexity will not work because it is only defined up to a constant.)

That being said, here is my preferred answer: 58, 26, 16, 14, 10, 666, 666, 666, 666, 666,...

The pattern is that Satan put the first five numbers there to mislead clever people and the rest of them are 666.

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    $\begingroup$ This is the most predictable response to such questions. But suppose you are taking a multiple choice test (I won't invent numbers here, but you get the point). Will you attempt to determine what the test authors think is the best solution, or leave it blank? $\endgroup$ Oct 18, 2012 at 19:17
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    $\begingroup$ I do agree that this is not math... $\endgroup$ Oct 18, 2012 at 19:17
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    $\begingroup$ @TheChaz I would write an answer in the margin like the one I gave above and expect it to receive full credit. $\endgroup$ Oct 18, 2012 at 21:11
  • $\begingroup$ Not much of a margin on scantron tests... $\endgroup$ Oct 18, 2012 at 21:43
  • $\begingroup$ @TheChaz I would still expect that if a multiple choice question were defective (e.g., if it did not have a unique correct answer) then it would not be counted toward the score on the exam. $\endgroup$ Oct 18, 2012 at 22:24
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I get $6$ for the fifth term:

$T(n)=\frac{378-272n+75n^2-7n^3}{3}$ gives the general term.

$58=T(1)$ $26=T(2)$ $16=T(3)$ $14=T(4)$ $6=T(5)$

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  • $\begingroup$ How did you come up with the equation? $\endgroup$
    – zz20s
    Jun 6, 2016 at 3:38
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    $\begingroup$ Polynomial interpolation is the first thing that came to my mind when I saw the question, glad to see that someone share my sense of humor. It's rather surprising by the way, normally I'd expect something like 10,356 or some random large number for $T(5)$. $\endgroup$
    – BigbearZzz
    Jun 6, 2016 at 4:07
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This has a very simple arithmetical interpretation as casting out nines from $\rm\:2^k n,\:$ performed by interleaved casting and doubling, i.e. repeatedly: sum the digits of $\rm\,n\,$ then double. Using this to compute $\:2^4\cdot 58\pmod 9$ we obtain the following computation

$$\begin{eqnarray}\rm cast &\rm out&\rm 9'\!s\quad\ double\qquad\\ \hline \end{eqnarray} $$ $$\begin{eqnarray} 5 + 8 &=& 13,\quad 2&\cdot& 13 &\,=\,& \color{#0A0}{26} \\ \color{#0A0}2 + \color{#0A0}6 & = &\ \ 8,\quad 2&\cdot&\ \ 8 &\,=\,& \color{#C00}{16} \\ \color{#C00}1 + \color{#C00}6 &=&\ \ 7,\quad 2&\cdot&\ \ 7 &\,=\,&\color{blue}{ 14} \\ \color{blue}1 + \color{blue}4 &=&\ \ 5,\quad 2&\cdot&\ \ 5 &\,=\,& 10 \end{eqnarray}$$

Hence $\: 2^4\cdot 58\equiv 10 \pmod 9.\ $ Notice that $\,58,\,\color{#0A0}{26},\,\color{#C00}{16},\,\color{blue}{14},10,\ $ the sequence of intermediate results above, is precisely the original sequence.

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The Answer is pretty simple once you notice it.

58, 26, 16, 14 ....

  • (5+8)*2=26
  • (2+6)*2=16
  • (1+6)*2=14
  • (1+4)*2=10
  • (1+0)*2=2

So just take the first number which is 58 and change it to 5+8, then multiply by 2 to get the next number from which you just follow the same steps, adding the two numbers together then multiply by 2.

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I got 13.5

2^5 = 32 2^3 = 8 2^1 = 2 2^(-1) = 1/2

58 - 2^5 = 26 26 - 2^3 = 16 16 - 2^1 = 14 14 - 2^(-1) = 13.5

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