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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?

Martin Sleziak
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jduncanator
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6 Answers6

49

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

Matt E
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  • i am still confused?!? can you please explain in more details thanks. .... –  Oct 17 '12 at 06:18
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    $5+8=13$ 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. – hpesoj626 Oct 17 '12 at 06:24
  • Amazing sequence... ! – mCasamento Oct 17 '12 at 08:40
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    not sure what this has to do with intelligence, though! – wim Oct 17 '12 at 11:19
  • @wim I take a VCE psychology class, and by Gardners theory of Multiple Intelligences one of your "areas" of intelligences is "Maths/Logic". The IQ test I'm guessing was designed around Gardner. – jduncanator Oct 17 '12 at 20:47
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    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. – Andreas Blass Oct 11 '13 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.

Trevor Wilson
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    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? – The Chaz 2.0 Oct 18 '12 at 19:17
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    I do agree that this is not math... – The Chaz 2.0 Oct 18 '12 at 19:17
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    @TheChaz I would write an answer in the margin like the one I gave above and expect it to receive full credit. – Trevor Wilson Oct 18 '12 at 21:11
  • Not much of a margin on scantron tests... – The Chaz 2.0 Oct 18 '12 at 21:43
  • @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. – Trevor Wilson Oct 18 '12 at 22:24
  • This is certainly a question *related* to mathematics. I don't see any reason to restrict this site only to questions that can be formally defined. – Tanner Swett Nov 01 '12 at 02:39
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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.

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

Guest
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  • How did you come up with the equation? – zz20s Jun 06 '16 at 03:38
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    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 $T(5)$. – BigbearZzz Jun 06 '16 at 04:07
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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.

Bill Dubuque
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1

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.

mau
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cameron
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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

Crooks
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