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Given options are:

  • A. 13
  • B. 42
  • C. 18
  • D. 30
  • $\begingroup$ Is this a general IQ question more then a mathematics? $\endgroup$ – Anurag Jun 28 '17 at 10:14
  • $\begingroup$ I assume 42 is good fit. Middle row (4 power 3 =64) and Multiple values of 1st row appear in below. (6*4=24, 6*8=48 & 6*7=42) $\endgroup$ – CR241 Jun 30 '17 at 21:38
  • $\begingroup$ B. The answer is 42. $\endgroup$ – Glorfindel Jul 9 '17 at 8:34

This is a long shot, and probably not the correct answer but seems to fit.

If you take the number of enclosed areas in the upper row and multiply by 4, you get the second digit in the lower row. If you take the number of "corners" or "loose" ends (I know -- this is a very vague definition :-D) and subtract 1, you get the first digit in the lower row.


6 has one enclosed area, three has three "corners" or "loose" ends. $(3 - 1) , 1 \times 4 = (2, 4)$.


8 has two enclosed areas, 64 has five "corners" or "loose" ends. $(5-1), 2 \times 4 = (4,8)$.


7 has no enclosed areas, 4 has four "corners" or "loose" ends. $(4-1), 2 \times 0 = (3,0)$. This gives D.


The answer is:

B: 42


It's 2*3*7


Every number on the board is made up of three factors, 2, 3, & 7. Every number with a three is in the far left column or the bottom row (and there's only one 'three' in a number). The number of twos in a number sharply increases in the central column, I assume there's a pattern I haven't figured out. There has to be a seven in our mystery number because otherwise there's only one on the board.


If we assume the mystery number has to have a factor of a three and a seven in it, then the number has to be a multiple of 21, and only 42 fits. Further its placement suggests only one or zero "twos" in it which limits our choices to 21 and 42.


@Dark Matter makes a telling observation, but I disagree with their conclusion.

My answer would be:

C. 18

Here's why.

Each cell contains a value that can be factored into powers of 2, 3, and 7 - the telling observation @Dark Matter made. It seems safe to assume the final value will also fit this pattern. From this, we can drop choices A and D, as they do not factor to powers of 2, 3, and 7.

Let's factor the values in the grid:

Powers of 2,3,7 per cell

 1,1,0  3,0,0   0,0,1
 0,1,0  6,0,0   2,0,0
 1,3,0  4,1,0   ?,?,?
(For example, bottom middle is $2^4\times3^1\times7^0=16\times3\times1=48$, so 4,1,0.)

@Dark Matter suggests that

there shouldn't be only one value that is a factor of 7. I disagree - I think the lone 7 is intentional, and that it is the "origin" of concentric layers, each layer having a pattern in the factorization.

Upper right cell is 0,0,1.
Its three neighbors are all n,0,0 - pure powers of 2.
The remaining cells we can see are x,n,0 - x is any value, n>0

To fit the pattern, the last cell should also be x,n,0. Our options are:
B. 42 - factors to 2x3x7, or 1,1,1 - not our pattern.
C. 18 - factors to 2x3x3, or 1,2,0 - fits our pattern.

So we choose C. 18.


I could not answer this question if there were not answer options. My answer:



choose numbers from rows: (6 + 8) mod 7 = 7 mod 7; (3 + 64) mod 7 = 4 mod 7; (24 + 48) mod 7 = 30 mod 7;


may be we can look at it this way.. 6*3+6=24 6*7+6=48 6*4+6=30

D 30 is answer

  • $\begingroup$ What about the other numbers? How do you correspond 3 to 24, 7 to 48 and 4 to 30? $\endgroup$ – boboquack Nov 9 '17 at 20:54

The answer is:



36-24=12, which is twice 6;
64-48=16, which is twice 8; so
18-4=14, which is twice 7

  • $\begingroup$ Welcome to Puzzling.SE! Would you be able to translate this into English? I'm not currently sure how this answers the question. $\endgroup$ – F1Krazy Aug 24 '17 at 8:06
  • $\begingroup$ Unfortunately $\dfrac{24-3}{2}=10.5$, not $6$, since $36$ isn't on the grid, so your rule doesn't work. I am going to edit this into English, if you don't mind. $\endgroup$ – boboquack Aug 24 '17 at 8:33

protected by GentlePurpleRain Dec 18 '17 at 19:45

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