358

It is impossible. Proof: Let the $7\times 28$ area be painted with black and white squares in a checkerboard pattern. Every piece will cover $2$ black and $2$ white squares, except the T-piece, which covers $3$ of one color and $1$ of another. Since there are $7$ T-pieces, a tiling that uses every piece cannot cover the same number of black and white ...


57

Answer: Reasoning:


52

TLDR: I'll fill the board and prove that the solution is unique. First, let's start by: I'll paint those green: Let's repeat those steps a few more times, using orange, blue, red and purple, in precisely that order: Now, We can easily fill the topmost white squares by that reasoning. They can't be filled in any other way: Now, let's look at: And by ...


25

This works! (I think) (Hopefully that’s clear enough how the shapes go) I got this mostly by thinking about how the blue and red can be placed such that the top and bottom of the green can be transported up and down without overlapping. Brilliant idea! Hope to see some more!


22

I think that this tiling is a valid Tetris stack:


20

The trick to this puzzle is to: (And here are those tilings: the center was already given, and the rest are obtainable from these by rotation.)


20

The X-pentomino tiles the plane, so that tiling is a good way to start. There are two ways to cut an 8x8 region out of that tiling. If one of the 4 central squares of the 8x8 region has an X centred on it, you get this or else you get this The latter can be easily improved by replacing the ones at the edges to give this A different way to get the same ...


20

I started with this: Pushed things this way and that, ended up with this: Similarly, on 9x9: And on 10x10: It took me a while to get there, but that one suggests an emerging pattern. And here is an expandable solution for any $2n\times 2n$ grid.


20

COMPLETED GRID The first step: Next: An important side note: Moving on: The top shaded region: Hopefully, finishing up:


19

How's this? I got this by square-izing the icosahedron graph:


19

FINITE PORTION OF ANSWER This can be extended infinitely in all directions - see my route to solving below for how. First a detour to explain how I made a tool (which competing answers could also use, perhaps improving on the finite number of differently-coloured tiles) by which ideas can be quickly tried using an Excel spreadsheet: Select all, and set ...


19

Here's an expandable solution for $n\ge 5$ (even or odd):


17

I haven't tried one of these before; I just stumbled across the question by accident. But I think I have an answer. You can build all 26 letters if you have a set containing: Image below:


16

Using:


15

Got the dissection part done. No idea about the native African word, though.


14

2x2 - area 108 - optimal 2x3 - area 72 - optimal 2x4 - area 108 - optimal 3x4 - area 84 - optimal 1x5 - area 54 - optimal 2x5 - area 304 - optimal 3x5 - area 576 - optimal 2x6 - area 240 - optimal 1x7 - area 1034 1x8 - area 432 - optimal 3x8 - area 5880 1x9 - area 585 - optimal Note that by subdividing the yellow rectangles: 2x3 indirectly solves ...


13

Challenge 1: Fit the four yellow birds into the tray, no overlapping. Rotation and reflection are allowed. Challenge 2: Fit the four yellow birds and the blue piece into the tray, no overlapping. Rotation and reflection are allowed. Challenge 3: Fit the four yellow birds and the red piece into the tray, no overlapping. Rotation and reflection are allowed. ...


13

I think the answer is using the following coloring: For other board sizes, Reasoning:


13



12

Here is a proof that $12$ is the smallest possible number of regions in any feasible solution. Consider an arbitrary division of an arbitrary rectangle into $n$ regions, such that every region has exactly five neighboring regions. We translate this picture into a so-called planar graph: each of the $n$ regions then translates into a vertex/point, and there ...


12

Finally arrived at this solution after playing around on https://www.scholastic.com/blueballiett/games/pentominoes_game.htm for way longer than I care to admit!!!! :) First pair (UI-TF) Second pair (WX-PY) Third pair (VZ-LN)


12

First, a generalizable solution for $1 \times n$, $n$ is even. By halving the rectangles, we can also obtain solutions for odd $n$, and the parts with just rectangles and no W-pentominos can be shortened. This is a way to tile This is optimal for this $a \times b$ because And here is a way to tile I've found two more, one for 1x4: and a rather large one ...


11

I found these solutions while playing around on https://www.scholastic.com/blueballiett/games/pentominoes_game.htm.


11

The answer is Here's the proof, step by step (EDIT: the result was right, but the specific argument had a hole; I think I've patched that now) :


11

I think the answer is this Explaining $16$ possible solutions Some notes on how the solution is obtained.


11

People have given some good upper bounds, how about a lower bound. However we can improve this ... However we can improve this ...


11

Here's another proof of the lower bound in Sriotchilism O'Zaic's answer.


11

Partial answer, Tasks 2-4 Finished Task 2 Finished Task 3 (what a doozy) Finished Task 4 (easier than 3)


11

The completed grid: Reasoning: Next: The bottom right: Closing in: Finally:


11

Some initial deductions: In the bottom right, Now, note And now, there's not much progress that can be made without thinking more globally. Continuing with this newfound knowledge, Finishing it off: The final answer:


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