# Wow, Puzzle. Woah, Numbers

I think it's either:

7. Multiply the two numbers on the right side. Ex: 7 x 3 = 21(top right ) 2 x 3= 6 (bottom right) 9 x 6= 54(top left) 4 x 6 =24(bottom left).

If the grid is on top, you subtract the numbers on the left side so it will create number that is divisible to the number we produced on the right. If the grid if on the bottom, the you add the numbers on the left side so it will be divisible to the number we produced on the right.

4-1=3. 3 is divisible to 21(top right) 3-1=2. 2 is divisible to 54(top left) 2+4= 6. 6 is divisible to 6(bottom right) 6+0=6. 6 is divisible to 24(bottom left)

Middle grid

we'll assume that the middle grid follows the same rule as the bottom grids. 4+3=7. So that means the question mark has to be 7 to make the rule possible. 7 x 3= 21. 7 can be divided into 21

I came here today with the hope that you guys would have a much better reasoning.

This puzzle was dusted off from the "very" boujee dating site known as iqcatch.com

• You should mention the sources of the puzzle. Please edit your post accordingly. Sep 15 '16 at 6:32
• It's from a intellectuals dating site i think? Ever heard of iqcatch.com? Lmao me either Sep 15 '16 at 6:34
• "Much unknown. Very arbitrary." - Doge Sep 15 '16 at 16:00
• I was more commenting on the puzzle itself (as well as an attempt at humor based on the title). The "find the pattern" puzzles from IQ tests so often because "find my pattern" puzzles, instead. Sep 15 '16 at 19:38
• @Engineertoast tbh, i didn't expect anybody here to get that reference. Sep 15 '16 at 19:53

2, indeed.

But the reasoning is different:

Start with the top left box, and move around the other boxes in a clockwise spiral towards the centre. Read each box as a pair of 2 digit numbers, one above the other. The sum of these 2 digit numbers follows the sequence 55, 60, 65, 70 and 75, box after box.
For example, the first box is $39+16$ then it is $47+13$, until the last box : $43+3?$, hence $?=2$.

• Very good solution. I think it would even better pointing out that the sum of the two numbers in the central box could not start the sequence in a clockwise spiral (50) so it has to be 75. Sep 15 '16 at 9:26

Is it:

2. Add all the numbers in a grid, add $1$, and take floor(sum/$5$), this number goes into the centre grid.

or, an alternative method:

Add all the number in a grid, and take floor(sum/$6$)$+1$, this number goes into the centre grid.

I think it is:

1

Because:

When you look at the other boxes, all of them have 2 numbers which when added together form one of the other (always the hightest) numbers. When we do the same to the box in the middle, the highest there is 4, so to achieve this, we need one to be added to three.

• By that logic it could also be 7 Sep 16 '16 at 8:22
• Oh shoot, you are right. Come to think of it, by the same logic it could also be 6 (3x3)... Sep 16 '16 at 8:23