In following sequence, Find the next number?
85, 79, 73, 69, ?
-7
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closed as unclear what you're asking by JonMark Perry, Marius, Engineer Toast, manshu, Gamow Aug 3 '16 at 15:45
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3$\begingroup$ can you give bigger sequence , difficult to guess with just 4 nos.. $\endgroup$ – Amruth A Aug 3 '16 at 8:38
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6$\begingroup$ On Puzzling SE we discourage 'find the next number in the sequence' questions as they generally are of lower quality than other questions. Practically any number can be justified as the answer — creating a polynomial is one of the many ways to do this. $\endgroup$ – Shuri2060 Aug 3 '16 at 9:34
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$\begingroup$ Not so much [logical-deduction] as it is [lateral-thinking], eh? $\endgroup$ – can-ned_food Mar 21 '17 at 10:42
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The values are in the reverse order of ASCII values of the Vowels
85, 79, 73, 69 equals to
U, O, I, E
So A's equivalent ASCII code is $65$
One more possible
The difference of 85 - 79 = 6
The difference of 79 - 73 = 6
The difference of 73 - 69 = 4
The difference of 69 - x = 4
So, x = 69 - 4 => 65
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3$\begingroup$ the answer did not mention ascii coding or anything with vowels. it was just a random guess, accidentally meeting the solution that you thought of. $\endgroup$ – elias Aug 3 '16 at 8:46
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$\begingroup$ @elias meta.puzzling.stackexchange.com/questions/4518/… $\endgroup$ – lois6b Aug 3 '16 at 8:47
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1$\begingroup$ @lois6b, fine, but this won't help the quality of the puzzles on the site $\endgroup$ – elias Aug 3 '16 at 8:54
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Totally random answer to the seemingly totally random question:
67
Because
subtracting 26 from each of those, results in consecutive primes in a reversed order: 59, 53, 47, 43. The next one in this sequence is 41, so adding the 26 back again, the final answer is 67.
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1$\begingroup$ i didnt downvote, but, i guess is because you are assuming the question is random ... $\endgroup$ – lois6b Aug 3 '16 at 8:45
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3$\begingroup$ I wanted to demonstrate that the question in its current form is too broad. You know the usual mantra about polynomials. $\endgroup$ – elias Aug 3 '16 at 8:51
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$\begingroup$ Usually the polynomial argument is kind of flawed - a number sequence can be a good puzzle if it has a nice / unexpected / interesting solution, and this answer you have given is much more pleasing than those ugly "roots of this polynomial" (with as many terms as given values of the sequence, where each term is multiplied by a 7 significant figure number). $\endgroup$ – Jonathan Allan Aug 3 '16 at 9:26
