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26, 20, 4, 16, 10, 2, 14, ?

I have tried everything.. I don't know how to solve this..

If you solve it post the solution please. Thanks!

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  • $\begingroup$ By searching this sequence, it looks like the answer is 8, with no explanation given. $\endgroup$
    – f''
    Commented Jun 25, 2016 at 18:36
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    $\begingroup$ Given how the other problems are on the test f'' found, I would not be surprised if it's supposed to be -6, /5, +12, -6, /5, ... $\endgroup$
    – Will
    Commented Jun 25, 2016 at 19:26

4 Answers 4

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Given $N_1=26,~~ N_2=20,~~ N_3=4,~~ N_4=16,~~ N_5=10,~~ N_6=2,~$ and $N_7=14,$ and working backwards from

$N_8=8,$

which f'' found somewhere, I can reverse-engineer

\begin{align}N_1&=26~~\text{(seed value)}&N_4&=N_3+12~~~&N_7&=N_6+12\\N_2&=N_1-6&N_5&=N_4-6&N_8&=N_7-6\\N_3&=N_2-2^4&N_6&=N_5-2^3&…\end{align}

… but that isn’t very satisfying.  You could probably construct equally plausible derivations for lots of other answers.

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  • 7
    $\begingroup$ This isn't a pattern and absolutely not satisfying. $\endgroup$
    – newzad
    Commented Jun 25, 2016 at 19:11
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I think it's

8

Because

          26
26 -  6 = 20
20 - 16 =  4
4  + 12 = 16
16 -  6 = 10
10 -  8 =  2
2  + 12 = 14
------------
14 -  6 =  8 (N_8)
------------
8  -  4 =  4 (the rest is hypothetical)
4  + 12 = 16
16 -  6 = 10
10 -  2 =  8
8  + 12 = 16

The pattern seems to be thus:

  1. -6, -16, +12,
  2. -6, previous (-16) divided by 2 = -8, +12,
  3. -6, previous (-8) divided by 2 = -4, +12, ...
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  • $\begingroup$ @Peregrine Rook, for what it's worth, I upvoted your answer. $\endgroup$
    – jonmsawyer
    Commented Jun 28, 2016 at 1:36
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Besides Will's comment another answer could be

0

Because if you

Multiply first digit with $x$ and second digit with $y$ you get the following sequence: $$\left\{2x+6y, 2x, 4y, x+6y, x, 2y, x+4y \right\} $$The sequence of the differences between each term is $$\left\{6y, 2x-4y, -x-2y, 6y, x-2y, -x-2y,\right\}$$If second term $(2x-4y)$ and 5th $(x-2y)$ term are equal then certainly there is a pattern. If they are equal then $x=2y$ and the transformed sequence become $$\left\{10y, 4y, 4y, 8y, 2y, 2y, 6y \right\}$$ and the sequence of differences become $$\left\{6y, 0, -4y, 6y, 0, -4y,\right\}$$Hence the next term (answer) will be $6y$ less then the term before which makes it $6y-6y=0$

If you don't want to go in details just

multiply the first digit with $2$ and the second digit with $1$, sum it to transform the sequence. You will see the some pattern.

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I found a pattern of my own:

Lets make it as:

26(a1), 20(a2), 4(a3), 16(a4), 10(a5), 2(a6), 14(a7), ?(a8)

There are two different kind of sequences going in a single one.

That is sequence of a1 -> a2 -> a4 , a5 -> a7 -> a8

26 -> -6 -> 20 -> skip -> 16 -> -6 -> 10 -> skip -> 14 -> -6 -> 8

In between a3 and a6 go like;

4 -> -2 -> 2

Complete sequence could be:

26, 20, 4, 16, 10, 2, 14, 8 , 0

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  • $\begingroup$ Well, the question specifically asks for the next (eighth) number in the series, and as far as finding $\mathrm{a8}$ is concerned, you've just found the same thing I already posted: $\mathrm{a2=a1-6}$,  $\mathrm{a5=a4-6}$,  and so maybe $\mathrm{a8=a7-6}$. $\endgroup$
    – Peregrine Rook
    Commented Jun 27, 2016 at 3:31

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