Find number in place of question mark

Find number in place of question mark, numbers have been written according to a certain rule.

  • $\begingroup$ is there one definite answer or could there be more than one scenario? $\endgroup$ Aug 8, 2018 at 8:00

2 Answers 2


I think the answer is:


the pattern:

the sum of the first row = 38
the sum of the second row = 45

the sum of the first column = 55
the sum of the second column = 38

The simple pattern is to make the sum from rows and from columns have the same number. So the answer should make the sum of the third row is equal to 55 and the sum of the third column is equal to 45. The answer is 19

  • 1
    $\begingroup$ your solution makes sense to me. i was kinda thinking the same way. don't know if it is the intended answer or not. $\endgroup$ Aug 7, 2018 at 8:41

The answer is


Steps of Solution:

1. Calculate the Square Root of second rows, the result will be 5,2,4.
2. Now, first rows coming from this formula: 2*4=8 (multiplying R2C2 with R2C3), 5*4=20 (multiplying R2C1 with R2C3) and 5*2=10 (multiplying R2C1 with R2C2).
3. Now the third row will come as: (5+2+4)*2=22, (5+2)*2=14 and (2+4)*2=12.

  • 3
    $\begingroup$ What is the logic behind the third row? $\endgroup$
    – hexomino
    Aug 6, 2018 at 10:47
  • $\begingroup$ @hexomino : The logic behind 3rd row is as below: From Second row, we have data as 5,2,4 (square root of all the columns of that row). So for R3C1, we have to perform (5+2+4)*2, i,e, [ (R2C1+R2C2+R2C3)*R2C2 ], and will get 22. Now for R3C2, logic is (5+2)*2 i.e. (adding [R2C1 and R2C2] and multiplying with R2C2 ). So, the result will be 14. Now the last one, for R3C3, logic is (2+4)*2 , [adding (R2C2 and R2C3) and multiplying with R2C2. So, the result will be (2+4)*2 = 6*2 = 12. $\endgroup$
    – Syed Rizvi
    Aug 6, 2018 at 17:08
  • $\begingroup$ But why (2+4)*2 and not, for example, (5+4)*2 [R2C1+R2C3]*R2C2? I don't see why one choice is favoured, there is no obvious symmetry here. $\endgroup$
    – hexomino
    Aug 7, 2018 at 0:21
  • $\begingroup$ Yes, may be that will also be possible and also the solution provided by @malioboro i.e. 19. $\endgroup$
    – Syed Rizvi
    Aug 7, 2018 at 5:06
  • 1
    $\begingroup$ 13 is also possible if after square root you do R3C1 = R1C2 + R2C2 and R3C2 = R1C3 + R2C3 so R3C3 = R1C1 + R2C1 = 8 + 5 = 13 $\endgroup$
    – wolfram42
    Aug 7, 2018 at 6:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.