I'm looking at three sequences from the test at http://free.ultimaiq.net/nse.htm. I managed to answer 27 of the 30 questions, but I'm really curious about the other three.

Q6: 1001, 4524, 4299, 3984, ? -- here is something like 100 145 244 299 398 45+99+55+99 and the next one should be 398+65and+99 but its not

Q24: 50625, 1296, 8100, 3240, ? -- I only know that the answer is not an integer and there is a hint "think how you can arrive at an+2 given an+1 and an".

Q22: 38, 50, 22, 18, ?, ?

New contributor
IDK138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • $\begingroup$ Q22 has the property that if you count the 7-segment LED bits set, you get 12,11,10,9, but I can't see any reason behind the choices. $\endgroup$ – JMP Nov 8 at 18:40
  • $\begingroup$ Yes i try many combination's to solve this question and still but thank you for that 2Q. $\endgroup$ – IDK138 Nov 8 at 18:44


The answer is 4354, which is +45,+99,+55,+99,+45,+99, so 443,542 splits into the 4-digit 4354.



The power of $3$ stays constant, the powers of $2$ and $5$ are the average of the previous two, so the answer is $2^{2.5}3^45^{1.5}=5122.889809472774517838207541981$, although anything with 1dp or more seems to work.

  • $\begingroup$ This is pretty stupid. (I don't mean that your answer is wrong, I mean the question is stupid.) Quite aside from the gratuitous "frame shifting", this boils down to asking the solver to extrapolate the sequence 45, 55, ? which could perfectly well yield 65 (OP's guess), 66 (triangular numbers), 45 (your answer), and any number of other roughly equally plausible possibilities. [EDITED to add:] I'm referring to Q6. Q24 is more reasonable. $\endgroup$ – Gareth McCaughan Nov 8 at 18:06
  • $\begingroup$ Thank you JMP for help. $\endgroup$ – IDK138 Nov 8 at 18:26
  • $\begingroup$ @JMP The much simpler way to explain the solution for Q24 is that a_n = sqrt(a_n-1 * a_n-2) (i.e. the geometric mean). $\endgroup$ – someone Nov 9 at 13:18

Your Answer

IDK138 is a new contributor. Be nice, and check out our Code of Conduct.

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.