4
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ By searching this sequence, it looks like the answer is 8, with no explanation given. $\endgroup$
    – f''
    Jun 25, 2016 at 18:36
  • 6
    $\begingroup$ 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
    Jun 25, 2016 at 19:26

4 Answers 4

6
$\begingroup$

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.

$\endgroup$
1
  • 7
    $\begingroup$ This isn't a pattern and absolutely not satisfying. $\endgroup$
    – newzad
    Jun 25, 2016 at 19:11
3
$\begingroup$

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, ...
$\endgroup$
1
  • $\begingroup$ @Peregrine Rook, for what it's worth, I upvoted your answer. $\endgroup$
    – jonmsawyer
    Jun 28, 2016 at 1:36
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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 $\mathrm{a8=a7-6}$. $\endgroup$ Jun 27, 2016 at 3:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.