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The puzzle is as follows:

The figure from below represents a set of 12 squares joined forming a closed loop. By using only the numbers from 1 to 12 fill in the blanks.

The condition is that, without repeating any number. It must be fulfilled that the sum of the four values written along each side of the largest square is the same and the smallest possible. According to this consideration. What is the maximum value of the sum of the numbers that are written in the blueberry shaded squares?

a loop of 12 squares, the perimeter squares of a 4-by-4 grid, with one pair of opposite diagonal squares shaded blueberry

The alternatives given are as follows:

  1. 9
  2. 7
  3. 11
  4. 3

Does a way to solve this, without going in circles, exist? I've been trying to do it on my own but with not much success.

If you are confused, the requirement is that each group of four squares on each side of the greater square must have the same value.

While attempting this I figured that the numbers from 1 to 12 could be split in pairs having the same sum. Let's say 13, as 12+1, 11+2, and so on will follow that condition. But this didn't help much. I think the given condition forces each diagonal corner of the greater square to make the same sum, to balance the inner sum (the squares which go in the middle). I've also noticed that in order to get the least possible sum it might help to group in pairs the numbers which are on top, let's say 12, 11, and so on.

I tried plugging in numbers for more than 40 minutes and I got stuck.

Does a strategy exist for this, other than just putting numbers randomly? Please include the logic used in answers, and if similar logic could be used in a similar environment. A wordy solution would be really appreciated.

Since this was found in an older IQ test I assume that the intended approach would not use any sort of sophisticated algebra or so. So please try not to present an algebra based solution because I doubt that would be the way for resolving this puzzle.

Can someone help me here? Were my initial thoughts along the correct track?

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Short way to guess at the answer:

To get the smallest possible numbers, we would want the corners to contain 1, 2, 3 and 4. Assuming there's a way to do this, that would make the largest possible 3+4 = 7.
So my guess is #2. 7.

Proving that this works:

It is required that the sum of all four numbers on each side is the same. It is also required that each sum is the smallest possible. Because each corner is counted twice and each side is counted only once, to get the smallest possible number, you want the 4 smallest numbers in the corners. So they should be 1-4.
If I add all the numbers together and count each corner twice, I will get the total sum of the 4 sides, and dividing by 4 will give me what each side should total (assuming that it is possible to sum to this number).
So the total should be 1x2 + 2x2 + 3x2 + 4x2 + 5+6+7+8+9+10+11+12 = 88.
That means we will need each side to equal 22. Let's try it.
1 + 4 + 5 + 12
1 + 3 + 7 + 11
2 + 4 + 6 + 10
2 + 3 + 8 + 9
So now I have 4 groupings of numbers that each add up to 22 and no group containing both 3 and 4, so I'm confident my guess was correct.

Now you asked about strategy:

The strategy for the guess is fairly straight forward. The strategy for grouping the 4 sets of numbers was:
I want my four groups to start with
1 + 3 + _ + _
1 + 4 + _ + _
2 + 3 + _ + _
2 + 4 + _ + _
This guarantees that 3 and 4 can be in opposite corners.

I almost certainly want my two groups containing 1 to contain 11 and 12.
I can't group 1 + 3 + 12 and 1 + 4 + 11 because both would need 6, so I group them the other way.

The same will be true for 2 + 3 + 10 and 2 + 4 + 9 (both need 7), so again, group them the other way.

There are few enough numbers that once you start with this strategy, even if you make a couple mistakes you will still come to the answer relatively quickly.

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  • $\begingroup$ Thanks for your nice explanation, but I'm still stuck at part of your justification. It is the part where you all the numbers as 1+1+2+2+3+3+4+4+5+6+7+8+9+10+11+12=88. Why the numbers from 1 to 4 appear twice?. Shouldn't they account for only once when put in the corners?. My only guess is that you did it in that way in order to obtain the sum for each side of the square, and from then on you got the pairs. But is that what did you had in mind?. I'm already accepting your answer but it would be nice if you could add a clarification on why do they appear twice. $\endgroup$ – Chris Steinbeck Bell Feb 24 at 22:54
  • $\begingroup$ By the way. I'd like if you could explain why are we looking for the smallest numbers?. The condition in the puzzle is asking for attaining the smallest sum in all edges of the square not specifically the smallest numbers. Does it really matter this clue?. I found it misleading. Would swapping the choices I mean putting the 12,11,10,9 wouldn't be the way to go for this?, so that the sum is the least?. Can you please check this part?. I hope you could attend these doubts please. $\endgroup$ – Chris Steinbeck Bell Feb 25 at 5:02
  • $\begingroup$ OK. I'll elaborate on the double numbers. $\endgroup$ – Joel Rondeau Feb 25 at 18:16
  • $\begingroup$ See if my added explanation helps you understand why putting 1-4 in the corners allows for the smallest possible sum. $\endgroup$ – Joel Rondeau Feb 25 at 18:24

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