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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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, ... $\endgroup$– WillCommented Jun 25, 2016 at 19:26
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4 Answers
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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.
answered Jun 25, 2016 at 19:09
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7$\begingroup$ This isn't a pattern and absolutely not satisfying. $\endgroup$– newzadCommented 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
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14 - 6 = 8 (N_8)
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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:
-6,-16,+12,-6,previous (-16) divided by 2 = -8,+12,-6,previous (-8) divided by 2 = -4,+12, ...
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$\begingroup$ @Peregrine Rook, for what it's worth, I upvoted your answer. $\endgroup$ 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
answered Jun 26, 2016 at 8:49
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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 $\mathrm{a8=a7-6}$. $\endgroup$ Commented Jun 27, 2016 at 3:31