What are the basic/advanced strategies used to find the next number in series.

I know the simple ones such as addition, multiplication etc. But recently I came into a question that goes on something like 812, 819, 823, 835, 834, 851(Don't try to solve this, I changed some numbers and there is no sequence). This is to illustrate that the sequence goes up and goes down at 835 then goes back up. What strategies can you use to solve it?

These questions are for an aptitude test so you have about 60-90 secs to solve each question.

P.S. I can't really post the question that was asked as I might be disqualified from the job process if I post it.

  • 5
    $\begingroup$ "What are the basic/advanced strategies used to find the next number in series.?" - in practice, one usually checks the OEIS first. $\endgroup$ Oct 21, 2011 at 15:25
  • $\begingroup$ It doesn't find the sequence in OEIS. I have about 5 of these sequences that increases, decreases and increases again. There is no sequence in the table on the website. $\endgroup$
    – daking963
    Oct 21, 2011 at 15:36
  • 2
    $\begingroup$ Without some restriction on what the formula must look like, this problem isn't even well posed: there are infinitely many formulas that generate any given finite sequence (the Lagrange interpolating polynomial of that sequence, for instance) and a lot of times the "right" answer to such puzzles has little if anything to do with math proper (such as the famous 1, 11, 21, 1211, 111221, etc sequence). $\endgroup$
    – user7530
    Oct 21, 2011 at 16:14
  • $\begingroup$ As a simple example of what user7530 is talking about, what's the next number in this sequence (2, 4, 3), given that the fourth number stands in relation to the third as the second does to the first? Well, one could make a case for 5, since (2+2)=4, and (3+2)=5. One could also make a case for 6, since (2*2)=4, and (3*2)=6. One could also make a case for 9, since (2^2)=4, and (3^2)=9. $\endgroup$ Oct 21, 2011 at 16:49
  • $\begingroup$ @Doug: what is the continuation of 4,2,4 ? It's 1. Why? Because when you were asked this question, you've been in a test for getting accepted in a course for becoming a trainer of soccer ... ;-) $\endgroup$ Oct 21, 2011 at 17:19

6 Answers 6


The answer is 42. Always.

  • 1
    $\begingroup$ A more likely answer is 42!! (only one exclamation mark!) Compare "G.H.Hardy and the number 42!" (go.helms-net.de/math/divers/GHHArdyAndTheNumber42.htm) ;-) $\endgroup$ Oct 21, 2011 at 22:57
  • $\begingroup$ @Gottfried, hmmm... so the dear Gottfr... Godfrey was in applied maths, after all? Who would have guessed so?! (Only a momentary weakness from the Great man, I suppose.) $\endgroup$
    – Did
    Oct 23, 2011 at 6:18
  • $\begingroup$ well, I won't exclude the possibility that he was even involved with the computation of the fractal border of Norway, and definitely he had somehow a relation with the (true lords of the universe, the) mice/mouse thing as you may find out if you google for Hardy and mouse... That must have been somehow "undercover" though, but I found at least one link... ;-) $\endgroup$ Oct 23, 2011 at 9:18

This answer is meant only if the context of such a question is some quiz or test, not if it occurs of, say a truncation of some approximated infinite series.

First arithmetical simplicity, then "translation" of the sequence to some current common knowledgde. Well, "simplicity" is a term lacking sharp bounds, so some "least square error" may be with the optimal guess...
Arithmetically: equal difference of first or second order, equal quotient. Equal quotient after removing a constant. Do the numbers have some common property, say they are squares, cubes. Possibly they are obvious moduli of some other sequence. Then I would try common sequences like primes, fibonacci..., just what is present in mind.
If no obvious answer, look at the form of the elements. Do they have only one digit? Perhaps the beginning of a well known constant.

If nothing matches so far, look at OEIS. If still nothing, it becomes harder.

But if it is so hard, then -if we are in a test- maybe the numbers represent something else, they are some transformation (like date or time information) or are numerical values for some common social construct related to required knowledge in the framework, where the test-situation occurs. Politics (years of important incidents...1,9,39,8,5,45,23,5,?? answer:49; why? (german historical knowledge required, sorry ;-) )) Economics, sports, social events... just what is up currently.
A third class are sequences, where the terms are codes, for instances code letters of the alphabet.

If all that has no success, I'd try to figure out the polynomial solution which is always possible...

An example for a non-obvious guess (illustrating some real-math-problems):

Actually, in my heuristic studies of functional and series-relations it is a problem at the heart of many problems to find approximations to rational numbers, if a function with a so-far unknown characteristic is approximated using only finitely many terms of its powerseries. It is much arbitrary to interpret some number like 2.44269504089 (of which I can assume it is only an approximation). Even if I have "the next one" 3.08136898101 and also the next one 4.00278070716, still for each one many interpretations can be found.
But if some interpretations of the three numbers have the common element log(2) with some rational number and small error , and especially, if the log(2) component has consecutively increasing powers of log(2), like (3.08136898101-2.4426950408) . log(2)~ 0.442695040954 and (4.00278070716-3.08136898101) . log(2)~ 0.638673940116 but again (4.00278070716-3.08136898101) . log(2)^2 ~ 0.442695040954 then I guess the continuation as something "obvious" and begin to try to make a proof for it - because of the argument of "arithmetical simplicity" at the beginning of my post...


About your series:

Let $$ \begin{eqnarray*} \begin{split} P(x)&:= 812\frac{(x-2)(x-3)x-4)(x-5)(x-6)}{(1-2)(1-3)(1-4)(1-5)(1-6)}+819\frac{(x-1)(x-3)(x-4)(x-5)(x-6)}{(2-1)(2-3)(2-4)(2-5)(2-6)}\\ &+823\frac{(x-1)(x-2)(x-4)(x-5)(x-6)}{(3-1)(3-2)(3-4)(3-5)(3-6)} +835\frac{(x-1)(x-2)(x-3)(x-5)(x-6)}{(4-1)(4-2)(4-3)(4-5)(4-6)}\\ &+834\frac{(x-1)(x-2)(x-3)(x-4)(x-6)}{(5-1)(5-2)(5-3)(5-4)(5-6)}+851\frac{(x-1)(x-2)(x-3)(x-4)(x-5)}{(6-1)(6-2)(6-3)(6-4)(6-5)} \end{split} \end{eqnarray*} $$

Then $P(x)$ is a polynomial which fits your data. This method is called Lagrange interpolation method.

BUT if $f(x)$ is any function defined on the integers, then

$$Q(x)=P(x)+(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)f(x)$$ also fits your data.

Most of these problems are actually asking for the simplest formula which fits that data, but really is there such a mathematical thing as "simplest". Lets say that our sequence is $2,3,5$ ( I picked only 3 to keep things simple). Which one would you say that it is the simplest description?

  • $2^{n-1}+1$
  • $\frac{n^2-n+4}{2}$
  • the nth prime..

These aptitude or interview questions are explicitly designed to give the most information about your raw smartness while being as hard to study for as possible. In other words, they want the answer to your question to be "you have to be smart and you cannot significantly improve your score by studying." To the extent to which these stackexchange answers help you, the test writers have failed.


Don't know exactly but next number can be 845. If you consider it as two series as follows

812 = 8 -- First two no's i.e. 1 & 2 819 = 8 -- 9+1=10

823 = 8 -- Next two no's after 1 i.e 2 & 3 835 = 8 -- 3+5=8

834 = 8 -- Next Two no's after 2 i.e 3 & 4 851 = 8 -- 5+1=6

So we can see that left side numbers last digits are incrementing by 1 (i.e.1&2 then 2&3 then 3&4 then 4&5 ....) --Sorry cant explain exactly hop you will understand.

And for right side the total is reducing by 2 in each step...

So the next number could be 8 -- 4&5 i.e 845.


Hey its a comibnation of two series as 812 823 834. 845....(difference is 11) 819 835 851 867.....(difference 16)

812 819 823 835 834 851 845 867 so next no is 845


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .