Least amount of (n)x(n+1) tiles to perfectly cover a 7x7 floor with two irregular tiles

Question

What is the least amount, $\mathscr{L}$, of $(n)\times(n+1)$ tiles that can perfectly cover a $7\times 7$ floor of unit tiles,

$\hskip2in$,

while still using these two red tiles:

$\hskip2in$

$$\tiny\text{NOT TO SCALE}$$

These two red tiles can be rotated or reflected. They are special and are comprised of 3 and 4 unit tiles respectively. Just to be clear, you put these two red tiles on the grid, and then try to find the least amount, $\mathscr{L}$, of $(n)\times(n+1)$ tiles, which themselves can also be rotated/reflected, to cover the grid perfectly with no overlaps. So you're going to have a certain number of $(n)\times(n+1)$ tiles and the two irregular tiles to cover the floor. "Cover" doesn't necessarily mean "contain within" the grid, but the two irregular pieces, and any tile you generate, needs to cover the grid orthogonally to the sides of the grid.

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For an interesting wrinkle to the original problem post your answer for the least amount, $\mathscr{M}$, of $(n_i)\times(n_i+1)$ tiles for any number of $i=1,2,3,4,...$ which are contained in the $7\times7$ grid. $n_i=n_j$ for $i\neq j$ is allowed. That is, you can have $n_1=n_2=2$, for example.

So if you choose $n_1$ to be $2$, then you'd have your first tile as $(n_1)\times(n_1+1)=2\times3$, then your second tile could be $3\times4$ corresponding to an $n_2$ equal to $3$. In other words, any number of different sized $(n)\times(n+1)$ tiles.

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_{I hope this turns out to be fun and less confusing than my first question. If it's not clear, then leave a constructive comment, and I'll answer it.}

Answer 1

The minimum is

7 tiles of 2 by 3

Explanation

The tiles have to be 1x2=2, 2x3=6, 3x4=12, 5x6=30, or 6x7=42, and 6x7 is impossible by inspection. They have to fill an area of 7x7-(3+4) = 42 tiles. 42 is not divisible by 12 or 30, so the rectangles have to be 1x2 or 2x3. Since the goal is to minimize the number of rectangles, 2x3 would be ideal if possible, which it is:

Answer 2

Just to be annoying and pedantic, I've done it with

3 6x5 tiles.

As seen here, the grid is covered perfectly with no overlaps.

Answer 3

You can cover the 7x7 square perfectly (with no overlap and no excess) with a 5x4, a 6x3 (4x3 + 2x3), two 2x1 and the prescribed irregular tiles, for a total of 5 n x (n+1) tiles.

Answer 4

Nobody seems to have answered part 2 yet, which is to tile the 7x7 with the given tetromino plus tromino plus all-different-sized n by n+1 rectangles. Here is a complete set of solutions of all the ways you can tile a 7x7 with the two given pieces plus any number of n by n+1 rectangles, allowing repeats. As you can see it is impossible. There are 97 100 440 tilings, with just those 28 sets of pieces.