Number-sequence puzzle where the number goes down in the middle

My friend asked me for help with an IQ test and after she did it online she came back at me with the ones she couldn't solve.

Two of them however, I discovered are the same puzzle type, but no matter how much I look at it, I just can't see the logic in it. We redid the test and this puzzle occurred yet again, so now I have three versions of the same type of sequence written below.

19 43 83 233 59 61 283 ?

11 35 75 225 51 53 275 ?

5 29 69 219 45 47 269 ?

I get the feeling that this puzzle is easy, yet I just can't see it.

[Update]

By request, I give the five options for the answer of the top row that I wrote down during the second test.

The options are: 800 778 793 58 176.

I'm actually leaning towards Jonathan Allan's explanation of a mistake in some data entry for their automated question generation right now, since no one here seems to have solved it yet. Prior to my post here, my friend emailed them asking them about this sequence, so most likely we will know between now and a few days.

[Update]

Looking at the options again, this puzzle is really really simple. Thank you smriti.

• Could you cite the source too please? Commented Aug 25, 2016 at 7:37
• Is it multiple choice question? Can you give options? Commented Aug 25, 2016 at 10:29
• Source is 123test.nl, I'll give the options when I get home. Commented Aug 25, 2016 at 11:16
• If all the choices except one are even, then I think I've got it. Commented Aug 25, 2016 at 16:54
• Ugh. Was the question posed in a way that makes this anything other than horrible? (E.g., was it "which of these numbers could come next?" rather than "What is the next number in this sequence?"?) Commented Aug 26, 2016 at 15:34

793: Why? They are all odd numbers! (Only possible to guess given the multiple choice nature of the question) The pattern means diddle squat.

None of the below matters, these were early attempts, which turned out to be wrong

In addition to @Daniel (Sorry, no rights to comment)

19 43 83 233 59 61 283 | 9 and 83

11 35 75 225 51 53 275 |1 and 75

05 29 69 219 45 47 269 | 5 and 69

then there is

19 43 83 233 59 61 283 | 59-19=40, 283-233=50

11 35 75 225 51 53 275 | 51-11=40, 275-225=50

05 29 69 219 45 47 269 | 45-05=40, 269-219=50

There are a few other simple additions with nice round numbers

d5-d1 = 40, d3-d2 = 40, d4-d2 = 140 (100+40)

d7-d4 = 50, d4-d3 = 150 (100+50)

d7-d3 = 200

Another pattern exists in the second digit... but this might be seeing patterns where there aren't any. It might also be giving even more credibility to the modulo + offset theory. (base+offset)%10 works for the patterns below.

9, 3, 1 -> 9, 1, 3 -> 9+2=11+2=13(drop the 10's)

1, 3, 5 -> 1+2=3+2=5

5, 9, 7 -> 5, 7, 9 -> 5+2=7+2=9

So:

If d1 + 40 = d5, and d4 + 50 = d7, then I will guess d6+60 = d8

logic: 5-1 = 4, 7-4 = 3, so 8-X = 2. X must be 6

61+60=121

53+60=113

47+60=107

• @Mithrandir - Possible conclusion Commented Aug 25, 2016 at 12:53

What these three have in common is that in all of these sequences the addition to the next is in this order:

24 40 150 -174 2 222 ?

So as these three sequences share the same changing sequence then the changing isn't related to the numbers themselves but the differences between the numbers.

Got some hints from @Shimizoki's reasoning, but this might be totally wrong:

N1 + 40 = N5

N4 + 50 = N7

This could mean that:

N7 + 60 = N8

Because in the next equation: The first in the equation is the number that is three positions further in the sequence. The second number increases with ten. The difference between the first in the equations place in the sequence and the sums number in the sequence increases by one less for each equation.

Therefore for the last number in the row for the first sequence would be 283+60=343.

This would mean the there are some decoy numbers and we have only deduced the pattern from two cases which makes me doubt that this is the right answer.

• Hello, and welcome to Puzzling! Nice spot for your first answer :) Commented Aug 25, 2016 at 7:23
• 5th-1st, 6th-2nd, 7th-3rd, 8th-4th = 40, 18, 200, ? hmm it doesn't really seem to be particularly logical as far as I can see. Maybe they made a mistake in some data entry for their automated question generation?! Commented Aug 25, 2016 at 7:34
• The sequence could contain a modulo function. Commented Aug 25, 2016 at 8:02
• Just some further observations: The first set are all primes. All sets are the same as each other, with an offset - also making all the sets primes with an offset: (-8 and -14 respectively), which might promote the modulo theory (@Meta45) Commented Aug 25, 2016 at 9:20
• I agree with those observations — nice spot. However, I wonder if this question really is an IQ question — modulo arithmetic isn't exactly elementary maths, so it seems to be more of a test of knowledge than IQ. Commented Aug 25, 2016 at 10:09

Using the first sequence and repeadetly finding (negative) differences:

19    43    83     233     59     61     283 / -3929
24   40    150     -174    2      222/-4212
16   110    -324     176   220/-4434
94    -434    500     44/-4654
-528    934    -456/-4698
1462   -1390/-4242
-2852/-2852


I added numbers after the slash based off the bottom.

Then, I offset it by 8 and 14 for the other patterns, so I get

-3929 -3937 -3943

• This is equivalent to fitting a 6th degree polynomial to the 7 data points we have from the sequence, which can be done to any sequence of length 7.
– JiK
Commented Aug 25, 2016 at 12:58
• Good answer. Also a valid point by JiK. Commented Aug 25, 2016 at 13:21

How about we try to normalize the numbers by removing the first from all subsequent. Hereby we get

0 24 64 214 40 42 264

Every number has a 4 in it, with many 2's and 6's.

Work in progress for others to consider. Is there a pattern to the numbers now?

• The first half of your answer is exactly equivalent to what MasterOfMuppets said about six hours before you.  But the second contains an insight that nobody else seems to have made, and which appears to be consistent with the correct answer. (Unfortunately, you haven’t found enough to predict the answer; but it’s not clear yet whether that it possible.) Commented Aug 26, 2016 at 7:13
• No.. Theirs was looking at how much it changed between each of the them (Delta of A to B, Delta B to C, Delta C to D). Mine looked at the fact that by removing the first as an offset of the rest, you return values that strangely EACH have a FOUR in them. Commented Aug 26, 2016 at 11:29
• The fact that $N_2-N_1,~N_3-N_1,~N_4-N_1,~\dots$, is the same ($24,~64,~214,~40,~42,~264$) for each of the three sequences — which I refer to as “the first half of your answer” — is mathematically equivalent to MasterOfMuppets’s discovery.  I acknowledged the fact that the second half of your answer — every number in your delta sequence contains a “4” — contains an insight that nobody else made. And, BTW, one of your upvotes is from me. Commented Aug 26, 2016 at 18:04

I interpreted it a different way, with each sequence as two strings with a prime as an index. It makes me think it's the same sequence that's been modified by the first number:

19
43  83  233
59  61  283
+16 -22 +50

11
35  75  225
51  53  275
+16 -22 +50

5
29  69  219
45  47  269
+16 -22 +50


Multiplying the 2 digit pairs and brings you products unusually close to each other(<100 in each case), but I didn't get any further than that.

Also Not sure how it helps, but those sequences are each unfairly weighted to a single digit, in the same positions.

19 43 83 233 59 61 283 | 3

11 35 75 225 51 53 275 | 5

05 29 69 219 45 47 269 | 9

I think numbers should be 111,103,97 respectively.

We take difference only those number whose last digit is same.

• But how can you deduce from n1+40=n5, n2+40=n3, n4+50=n7 that n6+50 must then be n8??? Commented Aug 25, 2016 at 13:31
• I never said that .....its a difference between them.... Commented Aug 25, 2016 at 13:33
• Then what is your reasoning? I can't get a grasp of it. Commented Aug 25, 2016 at 13:34
• There are total 8 numbers...If we pair all the numbers..then there would be difference between them is 40 ,50....in my answer there are two 40,and two 50 Commented Aug 25, 2016 at 13:36
• Still no reason to pair them like you do to get those differences in the first place. Sure we get nice and even numbers... Commented Aug 25, 2016 at 13:40

I don't know where it can lead, but (and for the first sequence only) :

(43 + 83) * 2 - 19 = 233

(59 + 61) * 2 + 43 = 283

Coincidence...? but this is not working for the 2 other sequences.

That was just to contribute a bit or give an idea to someone...

Above all that, either the scenario of the question is a reality... so nothing proves that the 3 sequences follow the same law... or it is a scenario to embed the question and, yes probably, the 3 sequences may follow one law...

• No, nothing proves that the three sequences follow the same law, but did you see MasterOfMuppets’s answer?  Go ahead and try to solve the puzzle on your own, but you should read the other answers before you post yours. Commented Aug 26, 2016 at 7:16
• @PeregrineRook, thanks... the point is I did not saw the update of the question before posting : ( ... before this update the possible choices were not known... that changes the problem a lot : ) Commented Aug 26, 2016 at 7:19
• And my point is that the three sequences have the same sequence of deltas — in other words, they are offset from each other by constants.   That seems to be a clear enough indication that the sequences are the same that any discovery that applies to only one of them is probably barking up the wrong tree.   (Then again, sometimes these things are trickier than they appear.) Commented Aug 26, 2016 at 7:25

The pattern is simply odd numbers. Of the choices, only one of them is an odd number.

75, 67, 61

$N_2-N_1=24\\N_3-N_1=64\\N_4-N_1=214$

Assume that each sequence is structured as an initial number followed by a succession of groups of three numbers: $$\{N_1\}\quad\{N_2,N_3,N_4\}\quad\{N_5,N_6,N_7\}\quad\dots$$

$N_5-N_2=16\\N_6-N_3=-12\\N_7-N_4=50$

So, for an arbitrary seed/starting number $x$, the sequence is \begin{array}l N_1=x\\[3ex] N_2=x+24&N_3=x+64&N_4=x+214\\[3ex] N_5=\,x+24+16&N_6=\,x+64-12&N_7=\,x+214+50\\ \phantom{N_5}=\:~~~N_2~~~+16\quad&\phantom{N_6}=\:~~~N_3~~~-12\quad& \phantom{N_7}=\:~~~~N_4~~~~+50\\[3ex] \llap{\text{assume }}N_8=\:~~~N_5~~~+16&N_9=\:~~~N_6~~~-12&N_{10}=~~~~N_7\:~~~+50&\dots \end{array}

And so the three sequences, extended to ten numbers, are \begin{array}{c|ccc|ccc|ccc} 19&43&83&233&59&61&283&\boxed{75}&39&333\dots\\ 11&35&75&225&51&53&275&\boxed{67}&31&325\dots\\ 5&29&69&219&45&47&269&\boxed{61}&25&319\dots \end{array}

• Welcome to Stack Exchange, and to Puzzling Stack Exchange.  We prefer answers that are clearer and have more explanation.  For example, it’s unclear when you use number dash number to indicate a range, since that is indistinguishable from number minus number — especially when the immediately preceding sentence is talking about subtraction, and the current sentence is a bit deficient in the verb department.  I’ve tried to clean up your presentation,  … (Cont’d) Commented Aug 26, 2016 at 7:15
• (Cont’d) … and it’s become apparent to me that you are conjecturing that $\{N_2,N_5,N_8,\dots\}$ forms a linear progression, and so do $\{N_3,N_6,N_9,\dots\}$ and $\{N_4,N_7,N_{10},\dots\}$.  Why?  That’s a very bold presumption, when all you have is the first two members of each of those subsequences.  It’s as if I said, “I have a sequence that begins $60,83,\dots$; what comes next?”, and you said $106$, because that’s the linear extrapolation of the numbers I gave you.  (My point is, any two numbers define a linear function.) You haven’t found a pattern; you’ve imposed one. Commented Aug 26, 2016 at 7:16