For 1, a $7 \times 7$ square can be tiled

Attachments:

starchan wrote: 1. Can a $7 \times 7~$ square be tiled with the two types of tiles shown in the figure? (Tiles can be rotated and reflected but cannot overlap or be broken) 2. Find the least number $N$ of tiles of type $A$ that must be used in the tiling of a $1011 \times 1011$ square. Give an example of a tiling that contains exactly $N$ tiles of type $A$. [asy][asy] size(4cm, 0); pair a = (-10,0), b = (0, 0), c = (10, 0), d = (20, 0), e = (20, 10), f = (10, 10), g = (0, 10), h = (0, 20), ii = (-10, 20), j = (-10, 10); draw(a--b--c--f--g--h--ii--cycle); draw(g--b); draw(j--g); draw(f--c); draw((30, 0)--(30, 20)--(50,20)--(50,0)--cycle); draw((40,20)--(40,0)); draw((30,10)--(50,10)); label((0,0), "$(A)$", S); label((40,0), "$(B)$", S); [/asy][/asy] Proposed by Muralidharan Somasundaran Number the tile $(R_i, C_j)$ according to the number $x$ where $i+j \equiv x \pmod{4}$. Hence, a tile of type $B$ will always contain all of the four numbers $0, 1, 2, 3$ whereas a tile of type $A$ will always miss exactly one of $0, 1, 2, 3$ so nvm

The answer is $2023$. Color the board in this manner: 1 2 1 2 1 2......1 2 1 3 4 3 4 3 4.......3 4 3 1 2 1 2 1 2....... 1 2 1 . . . 1 2 1 2 1 2....... 1 2 1 Now it is easy to see that every square covers every color and that every L misses exactly one. So if $x$ miss $1$,$y$ miss $2$,$z$ miss $3$, $t$ miss $4$ and there are $a$ squares we have the following system of equations $y+z+t+a=506^2$ $x+z+t+a=505\cdot506$ $x+y+t+a=505\cdot506$ $x+z+z+a=506^2$ So we have $(x,y,z,t)=(x,x+506,x+506,x+1011)$ and their sum is $4x+2023\ge2023$ and we are done And the construction is not hard but idk how to put pictures in aops(in principle you fill the first $3$ lines and the first $3$ columns with $2\cdot3$ and in the bottom right corner you put $2$ strcutures of $2\cdot3$ and the rest is filled with squares)

2. Every tile of type $A$ covers either $(1)$ two squares in an odd row and one in an even row, OR, $(2)$ two squares in an even row and one in an odd row. Let $m$ be the number of tiles of type $1$ and $n$ be the number of tiles of tile $2$. Since a type $B$ tile always covers two even and two odd squares, and odd rows combined contain $1011$ more squares than even rows, $m-n = 1011$. On the other hand, since each row has an odd number of squares, there must be one tile that covers only one square in that row. This tile must be of type $A$. Therefore $n\ge 506$. (The number of odd rows) So, the total number of tiles of type $(A) = (m+n) \ge 506 + 506 +1011 = 2023$. For the construction (or rather, a sketch for you to fill in), look at the top left $7\times 7$ square. Tile it as in part 1 of the problem. This requires $15$ tiles of type $A$. Now, for the remaining tiles in the three top rows and three leftmost columns, tile them using $2\times3$ boxes (each made up of $2$ tiles of type $A$). This takes $1004$ boxes, and therefore, $2008$ tiles of type $A$. So in total, $2023$ tiles of type $A$. Now, it is easy to see the remaining squares can be tiles using tiles of type $B$. Wish I could include a picture, but can't figure out how to do that from my phone.