I found this IQ test (http://free.ultimaiq.net/nse.htm) and can't seem to solve questions 18 and 19:

Q18: 120, 10, 3, ?, ?

Q19: 41, 23, 61, ?

  • 2
    $\begingroup$ Please don't delete a downvoted question and just repost it again. Instead consider whether there are ways to improve your original question. $\endgroup$
    – Stiv
    Nov 29 '21 at 14:14
  • $\begingroup$ It seems there are two distinct questions here, Q18 and Q19. Please ask them separately. $\endgroup$
    – Evargalo
    Nov 29 '21 at 14:23
  • 5
    $\begingroup$ @Evargalo I don't think we need twice as many IQ test number sequence questions. It can be fine to include two puzzles in one post. $\endgroup$ Nov 29 '21 at 17:10

Partial Answer


This looks like the sequence here written backwards.
In other words, the given elements correspond to the number of labeled cyclic subgroups of the symmetric groups $S_6, S_5$ and $S_4$, respectively, and the corresponding terms for $S_3$ and $S_2$ are 1 and 1.
Typing this into the site we find that this is indeed the correct answer.

Q19 (incorrect answer which I like)

If we reverse the digits of each element of the sequence, we get 14, 32, 16 which are the doubles of 7, 16, and 8, respectively. Looking at the face of a dartboard, we see that these numbers are clockwise adjacent and the next segment is 11 which, working backwards, would make the next number in the sequence 22

Unfortunately, this is incorrect and the correct answer actually is

The best reasoning I can come up with for this goes back to reversing digits and dividing by two again to give 7, 16 and 8.
If we take the cube of these numbers we get 343, 4096, 512 so the next entry would be the smallest number whose cube begins with 6 and that is 4.
Doubling and reversing indicates that the original missing entry should be 8.

I'm willing to concede that there may be better explanations.

  • $\begingroup$ Q19 - my first thought was to do with prime numbers as 23, 41 and 61 are all primes. So I tried to figure out if there was a logic jump pattern between the primes. Starting from 41, you can jump back 4 primes to 23, and jump forward 5 primes from 41 to 61 (or 9 primes from 23). But that's not a clear pattern $\endgroup$
    – Trent
    Dec 1 '21 at 0:38
  • $\begingroup$ @Trent also the answer as given by the site (ultimaiq) is not a prime but I agree, this could also have been a reasonable path. $\endgroup$
    – hexomino
    Dec 1 '21 at 9:55

Q19: 41, 23, 61, my thought is to multiply 41 * 3 and getting 123 then I take 123 and moltiply 12 * 3 getting 36 and then multiply 3 * 6 getting 18. the 1 of the nunmber 18 is already include in 61 so remains the digit 8. 41,23,61,8


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