-4

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How to solve q. 42

53 (3) 59 92 (4) 98 34 (2) 38 71 (?) 79

Thank you for your help

  • 2
    Note that while we can help, we like to see prior effort by the asker. That way, we don't feel like we're being used. – Frank Dec 1 '17 at 15:54
3

If you have numbers X, Y, and Z, it looks like the trick is to

multiply the second digit of X by Y then append the result to the first digit of X to get Z.
50 + (3 * 3) = 59
90 + (2 * 4) = 98
30 + (2 * 4) = 38

If this is the case, the answer is

? = 9

1

9.

Take the last digit of the numbers and divide the right number by the left one.

53 (3) 59 -> 3 9 -> 9 / 3 = 3  
92 (4) 98 -> 2 8 -> 8 / 2 = 4  
34 (2) 38 -> 4 8 -> 8 / 4 = 2  
71 (?) 79 -> 1 9 -> 9 / 1 = 9
  • Have you checked this answer? The riddle seems vague. As far as I see, 4 would work as well. There is a second pattern, where each middle number is half the difference between the two outside ones. (38-34=4, 4/2 = 2; 59-53 = 6, 6/2 = 3; etc) – JMac Dec 1 '17 at 15:46
  • @JMac your method doesn't work: 98 - 92 = 6 / 2 = 3, when it should be 4. Interacting with the numbers directly is something I usually do first. – dly Dec 1 '17 at 16:37
  • @dly I must have missed that one when I was checking them. I thought for a second that the question might have been bad enough to have multiple answers (and therefore without actually playing the game it would be hard to say which one it says is correct). Apparently doesn't matter though. – JMac Dec 1 '17 at 17:17

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