7
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A short number-riddle for warm-up:

There are 4 blocks with 5 numbers each. Which one fits in the place of the "?"?

22         4
     102
2           5


8           5
     264
11         6


6          11
    -711
2           3


9          8 
      ?
77        2

Edit: I rolled back the last edit, since the indention may be relevant for the solution ;)

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  • 1
    $\begingroup$ I hold back with up- or down-voting this puzzle until I see the answer. These puzzles can indeed be nice and sometimes beautiful, but we've had a lot of really bad examples here on Puzzling in the past. I hope it does not fit into that category. - Oh, and welcome to puzzlingSE! $\endgroup$ – BmyGuest May 1 '17 at 21:08
  • $\begingroup$ did you just change the question? $\endgroup$ – Oray May 20 '17 at 13:42
  • $\begingroup$ I'm voting to close this question as off-topic because he changed a number after days. $\endgroup$ – Oray May 20 '17 at 14:45
  • $\begingroup$ The 4 in the second block changed to a 5. Apparently there was an error, and one of the solvers found it below. He hasn't posted the solution yet though. Must be right if OP has changed the question though. It's a bit surprising how many up votes there are on this question. $\endgroup$ – Dr Xorile May 20 '17 at 14:48
3
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With these numbers it will be impossible to find the correct solution because there is a wrong number in block 2. 4 has to be 5.

The solution is now possible ;) Do you want me to post the solution?

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  • $\begingroup$ Can you please explain? $\endgroup$ – boboquack May 20 '17 at 1:05
  • $\begingroup$ You number in the middle of each block is calculated by the surrounding 4 numbers. The same calculation formula applies to each block. But with the wrong number in the second block this formula only applies for Block 1, 3 and 4. This has to be corrected and then the same formula applies to every block. You must not add additional numbers, you have to use the given numbers only. If you want to know the formula, i will post it here. $\endgroup$ – blauerlupo May 20 '17 at 11:32
  • $\begingroup$ Yes, of course ;) I am very sorry for the wrong number... $\endgroup$ – Ctx May 20 '17 at 13:27
  • $\begingroup$ The solution for the ? is: 328527 (22*5)+(2-4)^3=102 (8*6)+(11-5)^3=264 (6*3)+(2-11)^3=-711 (9*2)+(77-8)^3=328527 @Ctx: Where did you find this puzzle? Is it from a book? or somwhere from the internet? i am asking because this puzzle is shown on another homepage with the exact same mistake. $\endgroup$ – blauerlupo May 21 '17 at 16:00
  • $\begingroup$ @blauerlupo It is part of a bigger riddle on the net and I thought, it would fit well here $\endgroup$ – Ctx May 21 '17 at 17:38
5
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You may consider this question as;

$a*x+b*y+c*z+d*t=f(x,y,z,t)$

$a,b,c,d$ are the values given in the equation and $x,y,z,t$ are unknowns to solve $?$. So using wolframalpha you can find possible $x,y,z,t$ values which changes since there is not enough input as seen in the link. As a result, we can conclude that;

or more general, you can find different results for different $n$ values;

$9(-373n-222)+8(-236n-231)+77(-820n-480)+2 (2158n+1374)$

For example for $n=0$;

$-38058$

for $n = -1$;

$26011$

Actually you can find more answers by changing the equation (which fits with the given ones) as many different values as you want

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  • $\begingroup$ This is a generic solution on a pure mathematical basis, thank you! But there ought to be an unambiguous solution with the right idea. $\endgroup$ – Ctx Apr 27 '17 at 14:16
  • 4
    $\begingroup$ @Ctx And how are we supposed to know what you consider the 'right idea'? $\endgroup$ – jarnbjo Apr 27 '17 at 15:09
  • 9
    $\begingroup$ @jarnbjo Ok, when you have a series of m numbers you will always find a polynomial function f(x) of grade (m-1), which can fit this series for x in 0 .. (m-1) and claim, the next number is f(m). But usually, there is a more sophisticated pattern to detect, in many cases it is not even purely mathematical. It is the art to find this pattern. $\endgroup$ – Ctx Apr 27 '17 at 16:41
2
+100
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Still noone has an answer and puzzle looks hard, let me point out few small patterns, which I've found, may be they'll help someone else:

  • All two-digit numbers have same digits (11,22,77). So may be they aren't numbers, but digits written near each other with some different meaning.
  • The middle number is even when 3 out of 4 numbers are even and is odd when 2 out of 4 is odd.
  • Last digit of the middle number is always precent in the other 4 numbers. The first digit - is always not precent in the other 4.
  • Horizontal distance between last digit of each number is odd (11 or 9 cells). But it is even in the 4th block (10 and 8 cells).
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  • $\begingroup$ I'll award you the bounty manually despite you didn't solve it, since you showed effort. I will be giving a hint soon. $\endgroup$ – Ctx May 9 '17 at 11:15
  • $\begingroup$ @Ctx, thanks. Let me know when you add that hint, please:) $\endgroup$ – klm123 May 13 '17 at 9:45
0
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One possible pattern for the first block. 22 and 5 are farther away from each other than 2 and 4, and then (22*5)-(4*2)=110-8=102, which is in the middle. Not sure how to apply that further.

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-1
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Using the program Eureqa, we can find the following exact equation:

$y = ceil\Big(209 + 5x_3 + \frac{-x_2}{x_3^2}\Big) - 116\; mod(219 - x_3,\; x_2)$

$y$ is the center number, and $x_1$ through $x_4$ are the outer numbers reading top to bottom, left to right.

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  • $\begingroup$ The same problem here as with Oray's answer - Occam's razor - you take 4 numbers from nowhere. It would be easier if you just say: "The numbers in the middle should fit the following sequence: 102, 264, -711, 666. So the answer is 666". Even less assumption in this solution. And this is clearly now what the puzzle is about. $\endgroup$ – klm123 May 13 '17 at 9:47

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