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 '18 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 '18 at 6:56
  • $\begingroup$ Related joke $\endgroup$
    – A J
    Mar 27 '18 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 '18 at 9:22

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 '18 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 '18 at 12:41
  • 1
    $\begingroup$ maybe purple should start at 2424? $\endgroup$ Mar 27 '18 at 12:44
  • $\begingroup$ Looks plausible. $\endgroup$
    – user632
    Mar 27 '18 at 15:47
  • $\begingroup$ I think in the second sequence, there should be 1 more green term - 1221 $\endgroup$ Mar 18 '20 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 '18 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.