16
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Rule :

  1. enter image description here
  2. A cross sign means one rectangle's number is the multiple of another one.
  3. A line sign means one rectangle's number is adjacent to another one.

Example :

enter image description here

The Puzzle :

The numbers must be from 1 to 11, and each number can appear max 3 times.

enter image description here

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  • $\begingroup$ sorry for the missing rule $\endgroup$ – Jamal Senjaya Mar 13 '17 at 8:25
9
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The only possible solution is


11 is the number that is only used twice

The position of the three 1s is immediately clear from the minus signs. This helps a lot because we can now always disregard 1s for multiplication clues.

Next to the 5 there can only be a 2 or 3. If we use a 2 we would need at least four 3s to solve the grid (We would need another 3 below the 3 directly below the 1).

This means a 3 has to be placed next to the 5.

The left cell between the 6s can only have a 5 or 7, but a 7 would lead to a 1 next to it (and we already placed three of them).

Next we take a look left to the 5. A 6 would lead to a 5 and 11 which are impossible differences. This leaves 4 as the only possibility.

Above the top left 3 we need a multiple of 3 that has to appear another time. We already have three 3s and two 6s, so the only possibility is a 9.

We have to place a 4 or 8 on both sides of the 2 at the bottom right. We already used two 4s, so we have to place an 8 which leads to a solved grid.

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  • 1
    $\begingroup$ if that answer is correct, there seem to be two minus-signs missing between the top-most 10 and its adjacing nines. $\endgroup$ – beetkeeper Mar 13 '17 at 10:48
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    $\begingroup$ @beetkeeper Even though there are a lot of clues, it was never stated that all possible clues are shown. $\endgroup$ – w l Mar 13 '17 at 10:54
  • $\begingroup$ @beetkeeper : Thanks, you are right, I miss it. $\endgroup$ – Jamal Senjaya Mar 13 '17 at 12:24
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    $\begingroup$ @beetkeeper Also missing a minus for the starting 5 and the 6 to the right of it as well as the 5/6 in the top right. $\endgroup$ – David Starkey Mar 13 '17 at 13:29

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