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What will be the next number in the sequence?  66, 36, 18, ...  answer given in my book is 8 but how? Please explain. There does not seem any particular order of decrease.

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    There's no way to answer this. The sequence is far too short to say anything sensible about it. – lulu Aug 06 '17 at 15:58
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    As a quick example, if $p(n)=6n^2-48n+108=6\left(x^2-8x+18\right)$ then your sequence is $p(1),p(2),p(3)$. This would make the next term $p(4)=12$. Quadratic sequences are very simple...hard to see what's wrong with this answer – lulu Aug 06 '17 at 16:04
  • For the four values given ${8, 18, 36, 66, \cdots }{n=1}$ (reverse order as given) then sequence can be obtained from $$a{n} = 4 , n , \lfloor{ \frac{\lfloor{ \frac{n}{2} \rfloor} + 2^{n}}{n} \rfloor} + 1 + (-1)^{n}$$ – Leucippus Aug 06 '17 at 16:19
  • Related: https://math.stackexchange.com/questions/1790642/general-formula-for-the-1-5-19-65-211-sequence/1790666#1790666 – Jack D'Aurizio Aug 06 '17 at 19:21
  • The next number is clearly $801$, as the fourth number in the IP address https://www.tcpiputils.com/browse/ip-address/66.63.81.108, just in reverse. – Jack D'Aurizio Aug 06 '17 at 19:26

1 Answers1

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Let's start from 66 when we multiply two digits of 66 we get 36 which the next in the sequence .Similarly multiplying the digits of 36 we get 18 .Again multiplying the digits of 18 we get 8 which is the desired answer

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    That was probably the intended solution that OP's book had in mind. However, questions of this type ("what is the next number in this sequence?") are usually not specified enough, and this question is no exception. – Zubin Mukerjee Aug 06 '17 at 16:20