It is impossible.
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 ...
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:
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 ...
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.
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 ...
Finally arrived at this solution after playing around on
for way longer than I care to admit!!!! :)
First pair (UI-TF)
Second pair (WX-PY)
Third pair (VZ-LN)
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 ...
They are trying to prove that
How are they doing it?
Will they succeed?
A special case of the result being considered may be found at Maths SE (spoilers, obviously). The L-ish proof here uses the same underlying idea, but with the difference that
Since a single line doesn't do damage, it is possible to do
To achieve this, there are a couple of requirements:
To get these pieces one after the other
For this to occur so that the final piece also clears the board, we get these constraints on the number of pieces $X$:
Given these, the smallest $X$ that satisfies both requirements is
This is a small ...
I think I've found a solution for an 8x8 square. I do not know if it is the minimum solution or how to prove that:
It was definitely fun to try and find this! Took me a while. Excellent puzzle.
Some comments on how I got to the solution (Rather a chronology than a full deduction):
I believe I have found multiple solutions to the puzzle. It is possible that there is a nuance to the rules that I missed that disqualifies one or both of them. My understanding of the rules did not allow for a full logical deduction of the solution, so these were both found by brute force.
The sudoku is on the left, while the tetrominoes are on the right:
Here are a couple of $1 \times n$ solutions, which I believe to be optimal. They seem to follow some pattern(s), some of which are generalizable (spoiler ahead):
The solutions for $1 \times 10$, $1 \times 18$ and $1 \times 20$ (all below) also seem to form some kind of generalizable family.
As @JaapScherphuis notes in the comment, there's a (probably non-...
Here is my (now improved) solution for $2\times3$ rectangles plus N-pentomino(s):
And here is my solution for $2\times5$ rectangles plus N-pentomino(s):
Here are some more solutions, though for these I used computer assistance.
$1\times6$ rectangles plus N-pentomino(s):
The solutions given by others for $1\times3$ and $1\times5$ generalise to give ...
A few additions... firstly for the 1x5 a different answer, smaller area but more rectangles than Len's so just for interest
1 x 5 (Area = 160)
3 x 4 (Area = 400)
This one's the same area as Len's but fewer rectangles
1 x 8 (Area = 432)
Finally, a more interesting heptomino :) (in the sense that previous ones all had generalizable solutions who looked very much like this hexomino)
Here's the minimal solution for $1 \times 2$:
and for $2 \times 2$:
For $3 \times 5$:
My program found another one for $2 \times 7$:
a very narrow one for $1 \times 10$:
another one for $1 \times 11$: