15
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Fillomino rules (taken from Nikoli):

  • Fill in all empty cells with numbers under the following rules.
  • Divide all of the board into blocks. Fill each block with the same number horizontally or vertically.
  • Each block contains as many cells as the number in the block.
  • Same sized blocks cannot touch each other, horizontally or vertically.

Sum-until-1 variant rules:

  • Some numbers are written outside the grid. They indicate the sum of numbers into the grid from the position, until a 1 is encountered. The 1 is excluded from the sum.
  • Additionally, the number 1 appears exactly once on every row and column.

The following is an example Sum-until-1 Fillomino puzzle with its solution.

Now, solve the following puzzles "Odds", "Fibonacci", and "Pi":

You can solve online on Penpa: Odds, Fibonacci, Pi.

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Fibonacci was already answered correctly by trolley813, so just the other two:

Odds:
The placement of the 1s for this one is pretty straightforward. Starting from the bottom row, every time there is just one column where the 1 can go. After placing the 1s, everything basically falls into place without any difficult deductions.

enter image description here

Fibonacci:

Some easy deductions to start with based on the rule that the number 1 appears once in every row/column.

enter image description here

Now look at r2c3. If we expand the 5-block to there, we have two more cells to get to 13. Another 5 is not an option, since that would create a one cell area where a value of 3 is needed. Two 4s is not an option since we only have 3 cells available where those 4s could go. Therefore r2c3 is not a 5 which forces r1c4 and r2c4 to be 5.
The 8 on r5 can only be create by 4-4. Using 2 different values to get to 8 would require blocks of different size which is blocked by the 1 in r5c3.

enter image description here

The 2 in r3c5 can expand upwards or downwards. If it goes down to r2c4, the 4 and 5 block are forced and an additional 3-block is needed to get 13 in r2. This 3-block inevitable touches the already existing 3 block, which is not allowed. Therefore the 2 will have to expand upwards and with that the grid can be easily filled.

enter image description here

Pi:
This one is slightly more complicated, but again starting of with the 1s does the trick.
Start by noticing that the 5 area in the bottom right cannot extend left of row 4. If it would to get 15 on row 4 we could no longer have 5-5-5 since then there would be two connected 5-areas. 3-3-9 is not an option since there are not enough empty squares to fit 8 additional 9s. That leaves 4-4-7, requiring 7 squares for 7s, and 2 for 4s. Also we need an additional one for a one in column 1. That is 10 squares in total. If the 5s on the bottom right would extend beyond row 4, we only have 9.
This forces the 5s to go in r4c5 and r3c5 and therefore a 1 in the top right corner. With that the other two missing 1 are fixed as well. After this it is basically filling in the gaps again.

enter image description here

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  • $\begingroup$ I'm not sure I completely understand your logic for the Pi one. There seem to be some number/row/col mix-ups. Even with those corrected I don't think it explains why the 15 could not be made from numbers apart from three 5s. $\endgroup$
    – w l
    Jul 28 at 15:13
  • $\begingroup$ Thanks for the comment. There was indeed a row number mistake. I tried to clarify a bit more, but it feels overly complicated now. $\endgroup$ Jul 28 at 15:55
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    $\begingroup$ Welcome to PSE! The best received answers to grid-deduction puzzles are the ones that explain their logic step-wise, often with intermediate diagrams. Perhaps the best way to think of it is that not only are you trying to get "the answer", you're trying to provide a logical argument that your answer is unique. So while your explanation may seem overly complicated, it is exactly the sort of detail that well-received answers provide. I look forward to solving with you in the future! $\endgroup$ Jul 28 at 17:23
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    $\begingroup$ The new explanation is better, but I think it still needs to show why 4-5-6 is not possible (and 6-6-3, but that can get eliminated by the same logic you already used to remove the other possibilities). Could you also add the last puzzle, especially since the other answer is deleted now. That would make this a really great answer. $\endgroup$
    – w l
    Jul 30 at 9:05

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