5
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What number should replace the "?"

1 13 15
7 7 10
8 5 11
4 5 ?

Source: A modern approach to verbal and non-verbal reasoning - RS Aggarwal

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    $\begingroup$ Welcome to Puzzling! Is this a puzzle you created yourself? If not, please add a source (such as a link). We have an attribution policy here and unsourced puzzles will be closed and perhaps deleted. $\endgroup$
    – bobble
    Feb 18 at 5:56
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    $\begingroup$ Thank you bobble! I was not aware of the attribution policy, I have added the source now. $\endgroup$
    – ribbon
    Feb 18 at 6:12
  • $\begingroup$ I have taken my answer down because of the downvote $\endgroup$ Feb 18 at 12:47
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    $\begingroup$ @Smartest1here you were very quick about deleting the answer, so I'm adding the reason for downvoting the answer here instead: The puzzle, as given, has 11 numbers, arranged in rows and columns. The deleted answer depends on a property that ignores the rows, the columns, and an arbitrary pair of the eleven numbers. Actually, it is pretty much exactly equivalent to rearranging all the numbers in the grid, and then solving that grid. Since a complete rearrangement absolutely destroys any pattern in the original puzzle, any answer gained by such means can only be correct by coincidence. $\endgroup$
    – Bass
    Feb 18 at 13:20
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I think Answer will be

6 ??

Reason:

Sum of all first 3 numbers (A+B+C) in each column divide by 4th number (D) is result of last row. (A+B+C)/D

(1+7+8)/4 = 4
(13+7+5)/5 = 5
(15+10+11)/? = 6
It seems in sequence next number might be 6 (4,5,6)
Number Six (6) will be replace question mark..
(15+10+11) is 36.. 36/6 = 6

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    $\begingroup$ That looks very good! +1 $\endgroup$ Feb 22 at 22:24
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    $\begingroup$ @Albert.Lang thanks 😊 $\endgroup$
    – CR241
    Feb 22 at 22:41
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    $\begingroup$ Well done! I would say that this is almost certainly the intention. Alternatively, you could say that rot13(gur fhz bs gur svefg guerr ebjf vf gur fdhner bs gur sbhegu) $\endgroup$
    – hexomino
    Feb 23 at 10:17
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I think the answer might be

10

Reason

If we, instead of regarding the numbers numerical value, pay attention to how many letters they contain, we can see a pattern. The third column is a product representation of the absolute difference between column 1 and column 2. ONE(3) THIRTEEN(8) --> 8-3 = 5 = 1x5 (15). SEVEN(5) SEVEN(5) --> 5-5 = 0 = 1x0 (10). EIGHT(5) FIVE(4) --> 5-4 = 1 = 1x1 (11). And lastely FOUR(4) FIVE(4) --> 4-4 = 0 = 1x0 (10)

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  • $\begingroup$ Wouldn't 60 be just as valid a deduction? $\endgroup$
    – Bass
    Feb 18 at 20:12
  • $\begingroup$ I assume the puzzle creator didn't want to make it too obvious, thus added a "1" infront of the letter-difference $\endgroup$ Feb 18 at 20:23
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I'll throw out an obvious answer:

The missing number is 4.

The sum of the first column is 20, and the sum of the second column is 30. 4 is the number that makes the sum of the third column 40.

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    $\begingroup$ So, if you decide that the individual numbers and their actual position in the grid don't matter, then there's a way to combine them in a way that gives you two numbers that aren't the same as each other. Instead of revealing any kind of pattern, this seems more like a method you can trivially apply to any set of numbers, with equally convincing results. $\endgroup$
    – Bass
    Feb 18 at 20:10
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Two proposals:

1331, 14

Reasoning for 1st:

Number of distinct prime factors in each row: 3
Number of prime factors with multiplicity per row: 3,4,5,6
Why 11^3 and not some different prime number?
Product of distinct prime factors of columns 1 and 2 just below product of distinct prime factors in column 3 where largest prime factor in column 3 is as small as possible: 13 < 3x5, 7 < 2x5, 2x5 < 11, 2x5 < 11

Reasoning for 2nd:

Each number in column 3 is the sum of two numbers in the first two columns and not the same row as the sum, and each number is referenced exactly once.

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I think the answer is

19???

Columns:

If you look at the number of letters in the number itself, and add them all up in the column (Except the number in row 4, which is at the bottom, so add the number itself.) Keep doing this until i got these numbers:

13 and 19. These numbers make it so that the sum is one away from a multiple of 4. Pretty cool huh?

Rows:

You do the thing again and we still have the numbers: 13 and 19.

Diagonals:

This 4-sided table is not entirely square, so i should just do the 7, 5 and with the ? square. So i did it again, and something happened. When i did it by 13, i got 22 which is not one away from a multiple of four. BUT... when i did it by 19. I got 28 WHICH IS A MULTIPLE OF FOUR.

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The missing number is 14.

We observe all 3 columns have 8 odd numbers and 3 even. The sum of the odd numbers is

1+13+7+7+5+5+11+15=64. If we replace the question mark with the number 14 then the sum of

the even numbers is 4+8+10+14=36 and 64+36=100. The sum of the numbers from column 1 and 2

is 1+13+7+7=8+5+5+4=50 which is equal to the sum of the 3rd column 15+10+11+14=50

The sum of two numbers from columns 1 and 2 equals one number of column 3.

7+8=15

5+5=10

7+4=11

1+13=14

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  • $\begingroup$ You seem to be making a lot of arbitrary connections and you've also reached the same conclusion that's already been proposed here (and also in one deleted answer). This comment under the question nicely sums up why this is unlikely to be of value. $\endgroup$
    – El_Vanja
    Feb 26 at 16:42
  • $\begingroup$ @ El Vanja I did not look at your answer. My reasoning has considerable differences from yours.. $\endgroup$ Feb 26 at 18:18
  • $\begingroup$ It's not my answer. And even though you came to the same conclusion in a different way, it boils down to "The sum of two numbers from columns 1 and 2 equals one number of column 3." $\endgroup$
    – El_Vanja
    Feb 26 at 19:39
  • $\begingroup$ @ The main part of my answer is the square of odd numbers plus the square of even numbers is also a square. 10^2=8^2+6^2 $\endgroup$ Feb 26 at 22:17

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