This puzzle is part of the Monthly Topic Challenge #2: IQ Puzzle Parody

I have no idea how to approach this problem. It is from an IQ test. If you can give me some hints, I will be really thankful. If you give the full solution, I'll be extremely thankful. Please, help me pass this test.*

‣‣ Fill in the circles with the numbers given within the brackets to make a valid equation. Any number can be used more than once.

001. ( 2 + √〇 + 〇 ) × 〇 × ( 1 + 〇 ) = 210    (2,4,7)

010. 〇 ! × ( √〇 + 6 - 〇 + 25 ) - 〇5 = 1    (0,2,4,7)

011. ( 〇 + 4 + 2 - 1 ) × ( 〇 + 4 ) - 〇5 = 0    (0,1,2)


.. when formatting this problem, I had to copy and paste some symbols (circle, square-root symbol, ...) from the internet because I didn't have them in my keyboard. I don't want you to bother doing the same thing, so feel free to give me only the numbers without other symbols. But please give all the numbers in order. Otherwise I will have a hard time understanding what the correct numbers are.

Thank you in advance.


I didn't find the questions 002-009. They haven't provided them, I don't know why. (A typo?) But please don't worry about that. Ignore them just like I did.

* In case you are not aware of the Monthly Topic Challenge, I created this problem and I do know the solution : )

  • $\begingroup$ For 001: 210 = 2*3*5*7 (2+sq(4)+2)*7*(1+4) $\endgroup$
    – Hassan
    Commented Aug 20, 2022 at 21:29

2 Answers 2


Credit to @EdMurphy for the correct substitutions.

The question says to

write all the numbers, and only the numbers.

Thus we get:


Now these all have the same length, so I think I'm on the right track.

But now I'm stuck. Current ideas:

- The largest digit is 7, so the solution may involve converting from octal to binary/decimal/hex.
- All 12 of the digits in the above numbers have at least a pair (some triplets) of equal digits among the three numbers. The solution may involve choosing the "majority" number to create the single number 011042714250. Also the "minority" or difference could be used.
- The emphasis on the word "symbols" suggests that unicode might be involved, but all the supplied numbers lead to unused characters.
- Breaking the numbers into two-digit pairs might make for nicer results in (say) ASCII, but they don't all translate directly to the visible character region.
- There are quite a few funny symbols in the OP besides the ones used in the equations, including two triangle bullets (0x2023) and some non-break spaces (0xA0). There's also a 3-2-3 pattern of ellipsis which I suspect must be intentional.

Edit 1:

Per author's comment, the numbers we have are


Also, the question numbers

are clearly 1/2/3 in binary. The padded zeros also indicate that octal may be relevant. Therefore I think that the solution involves converting the digits from octal to binary.
Supporting this idea:
- 0, 1, 2, and 4 are more common than 3, 5, 6, and 7, indicating a binary distribution with more 0's than 1's. - We have 3 strings of numbers that are 9 long. If the numbers are expanded into binary, that could give us a 9x9 grid.

Exploring that last idea,

we have two natural ways of arranging the binary conversion of a number--vertically or horizontally. Testing each gives the following patterns: enter image description here These don't appear particularly useful, so I'm probably lost

That's all I've got for tonight; I might try some other stuff in the morning

Edit 2:

So I guess I lied in that last note, and tried one more thing.

Making the numbers into a 27x3 grid (ie. horizontal but no wrapping) we get the following pattern: enter image description here Reading the unset bits, we get the message in (low-res) Latin characters

Final Answer:


  • $\begingroup$ I think binary must have something to do with it, since the questions are numbered 1, 2, 3 in binary. $\endgroup$
    – Obie 2.0
    Commented Aug 21, 2022 at 5:24
  • $\begingroup$ OOPS.. I've forgotten one 'L' . What a shame. :( $\endgroup$
    – ACB
    Commented Aug 21, 2022 at 6:17
  • 4
    $\begingroup$ @ACB I prefer to think of it as a metalworking instruction step label :) $\endgroup$
    – ash4fun
    Commented Aug 21, 2022 at 6:29
  • 1
    $\begingroup$ @ACB I think it'd be best to leave the problem as-is. If you really want, maybe put the new equations as a footnote, but I would avoid changing/removing the original equations. $\endgroup$
    – ash4fun
    Commented Aug 21, 2022 at 16:12

These all involve simple deductions along the lines of "if this circle contains X, then this other circle must contain Y", and continuing until exactly one set of choices fails to lead to a contradiction. In light of the monthly theme, though, I feel like I must be overlooking some implied meta-puzzle...

Anyway, the solutions to the surface puzzles are:


4 2 7 4


0 4 7 2


0 1 2


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