33, 66, ?, 1212, ?, 1218

I just can't figure out how to complete this sequence. The last number throws off everything I try. Anyone have some ideas? I can perhaps see the missing numbers ending in 9 and 15, but that still might be the wrong pattern.

Taken from: https://news.generiq.net/Trilogica/algebrica.html

  • 2
    $\begingroup$ They make a few handful of these questions (32, to be exact), label it "IQ test" and is asking 35-40€ to grade the results? Good lord... $\endgroup$
    – votbear
    Mar 27, 2018 at 6:43
  • 15
    $\begingroup$ @votbear it's pretty simple, if you are dumb enough to pay, it means you have failed the test $\endgroup$
    – Kepotx
    Mar 27, 2018 at 6:56
  • $\begingroup$ Related joke $\endgroup$
    – A J
    Mar 27, 2018 at 7:08
  • 1
    $\begingroup$ If you got this from somewhere the most obvious answer is that the last number is a misprint. $\endgroup$
    – Willtech
    Mar 27, 2018 at 9:22

4 Answers 4


I think this is it.

Goes up in multiples of $3$ and number of each colour increases each time. $$\color{red}{33},\color{blue}{66},\color{blue}{69},\color{green}{1212},\color{green}{1215},\color{green}{1218},\color{purple}{1521},\color{purple}{1524},\color{purple}{1527},\color{purple}{1530}$$.


First digit(s) doubles each time. $$\color{red}{33},\color{blue}{66},\color{blue}{69},\color{green}{1212},\color{green}{1215},\color{green}{1218},\color{purple}{2424},\color{purple}{2427},\color{purple}{2430},\color{purple}{2433}$$

  • $\begingroup$ Would have upvoted if I didn't reach my daily limit :\ :D $\endgroup$
    – Mr Pie
    Mar 27, 2018 at 12:21
  • $\begingroup$ @TheSimpliFire If you are correct isn't there a gap between the blue and the green colour? $\endgroup$
    – rhsquared
    Mar 27, 2018 at 12:41
  • 1
    $\begingroup$ maybe purple should start at 2424? $\endgroup$ Mar 27, 2018 at 12:44
  • $\begingroup$ Looks plausible. $\endgroup$
    – user632
    Mar 27, 2018 at 15:47
  • $\begingroup$ I think in the second sequence, there should be 1 more green term - 1221 $\endgroup$ Mar 18, 2020 at 9:20

I'm new here. Another solution could be

129 and 615. Basically, we sum up the digits of previous number and append the consecutive multiple of 3 as follows:

3+3=6 (6),
6+6=12 (9),
1+2+9=12 (12),
1+2+1+2=6 (15),
6+1+5=12 (18)

That'd make the next one


  • $\begingroup$ Welcome to Puzzling.SE! It's customary here to hide answers inside spoiler tags (type >! at the start of each line), to avoid spoiling the solution for anyone who wants to have a go at the puzzle themselves. That's a pretty clever solution, btw! Feel free to take the tour and visit the help center to learn more about the site, and I hope you stick around! $\endgroup$
    – F1Krazy
    May 10, 2018 at 7:47

The complete sequence is:

33 66 39 1212 1515 1218

This because:

33=first number of sequence
66=add 3 to first digit and add 3 to second digit
39=first digit of first sequence number and add 3 to second digit of second sequence number...


33, 66, 69, 1212, 1215, 1218

Explanation (not sure though if it makes sense):

2nd digit: 3n

1st digit: 3n if log2 of n is an integer, and n-1 otherwise


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