5
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Which?

There's a 4 by 4 grid and a symbol in each box. Each symbol consists of a letter (A, B, C, or D) and a combination of a few shapes. Deduce which symbol is missing. Four options are presented.

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2
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One potential explanation:

Every letter must be at some point surrounded by a square and a circle (not necessarily at the same time).

The answer is:

Option 3 as B so far lacks a square

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  • 1
    $\begingroup$ Your solution is essentially that every letter must be in a square once-so most of the info in the grid is almost redundancy. Good observation but not the correct answer $\endgroup$ – Artemmm Apr 9 at 18:12
1
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Looks like:

The count of distinct things in a row/column adds up to 7 (so a different letter or style counts as 1 but only once in a row)

So if we apply that to the row and column our questionable tile is on:

Row 3 has D,A,circle,line and box - five distinct things. Column 1 has D,C,B,box and circle - again five distinct things.

And then:

We find that none of those answers fit (I think) because they're repeated. Option 1 gives no new options to the row and only adds one to the column. 2 adds 1 to row and none to column. 3 gives one to row and one to column. 4 gives one to row and one to column. Either way we don't get the extra two we need. The answer would have to be something like B,triangle, line for this to work. Ah well...back to the drawing board. I was so convinced every row and column adding to 7 was relevant...

____Old answer:________________

A potential rule:

We have letters and styles, each row/column either repeats a letter or repeats a style but won't repeat both.

Applying that:

Row 3 has two Ds and an A so we need a C or B without one of the other styles - The B in a box fits this. This also fits for column 1

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  • $\begingroup$ The last row contains distinct letters and no repeated styles. No cigar $\endgroup$ – Artemmm Apr 9 at 17:59
  • $\begingroup$ @ArtemLugin ah, drat, I thought I'd checked them all. Thanks for the spot. I'll keep thinking... $\endgroup$ – Lio Elbammalf Apr 9 at 18:04
  • $\begingroup$ for the new idea: row 4 contains all 4 letters+4 distinct symbols=8 $\endgroup$ – Artemmm Apr 9 at 20:53
  • $\begingroup$ @ArtemLugin It doesn't have a circle. It has A,B,C,D,line,square, triangle = 7 distinct objects (so we aren't counting square + triangle, square + line, square, line as four different things but breaking them up into their component parts and only counting each once. Every line does add to 7 so it could be a solution but there just isn't an answer to cover it. $\endgroup$ – Lio Elbammalf Apr 10 at 18:17
  • $\begingroup$ I see what you mean. But then the second column from the left adds to 6:a+b+d+sq+cr+ln. $\endgroup$ – Artemmm Apr 10 at 21:17
1
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I think the answer is

option 4: D inside a triangle.

Here is one possible way of approaching this puzzle:

Assign each letter and shape a number so that a slot is assigned a sum of letters and shapes. Figure out an assignment, so that filling a missing slot will turn the 4x4 square into kind of a semi-magic square: all the rows and all the columns will have same sum.

I made the following assignments

A = 8, B = 2, C = 4, D = -3

Circle = 3, Square = 1, Slash = -5, Triangle = 9

Option 1 becomes 12;
Option 2 becomes 7;
Option 3 becomes -2;
Option 4 becomes 6;

Now the square turns into:

7 9 -8 5
2 5 13 -7
6 -5 11 1
-2 4 -3 14

Option 4 is the only choice to make a puzzle into a semi-magic square.

I came up with this assignment by solving a very simple system of linear equations. The set of solutions is parameterized by letter B and a square shape :-)

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  • $\begingroup$ This is very interesting. Although not the correct solution. There's an infinity of ways of creating an integer magic square like this with any of the 4 options (correct solution has no such ambiguity). How did you find the nice numbers? $\endgroup$ – Artemmm Apr 10 at 2:55
  • $\begingroup$ I made a value of a letter and a shape a variable, then wrote a simple system of linear equations to satisfy semi-magic square rules. Solving this system gave me a very simple linear expression for the missing square. Another obvious constraint is that different letters should be assigned different numbers. The same goes for the shapes and the values for the options. In this parameterized solution options 1 and 2 did not satisfied these constraints. I went with option 4 (option 3 could have been also used for a different series of assignments) and picked some small values for parameters. $\endgroup$ – ppgdev Apr 10 at 4:04
  • $\begingroup$ @ArtemLugin, Actually I take it back. In my approach Option 3 would not have worked either. If I went with Option 3 I would have no choice but assign a circle and a slash the same value. So Option 4 is the only choice if you want to get to a semi-magic square and there is no ambiguity. Looks a like it is another "correct solution" ;-). $\endgroup$ – ppgdev Apr 10 at 4:19
  • $\begingroup$ To get a magic square like this is possible in an infinity of ways. As you're allowing repeated values (2 5s) some code can find many of them easily. It's not another correct solution but still an interesting observation $\endgroup$ – Artemmm Apr 10 at 9:20
  • $\begingroup$ D does stand out in that, like you said, the values you assign can all be distinct. To me its a strange thing to differentiate by as it has no influence on the magic squares themselves beyond calculation. If it were "D is the only option where magic squares or something else is possible" that would be another thing $\endgroup$ – Artemmm Apr 10 at 13:54

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