This problem looks like a traditional Saint-Petersburg or 239MO problem, respectively. It is kind of common usage to make use of such problems over and over again.

solution from Christian Sattler(Germany): first we define an involution on the hooks: send every hook to the one which includes the square where the hook goes around. it is clear that this is an involution, so the number of hooks is even. it immediately follows that 12 | mn. there are three cases(wlog 3 divides m): 3 divides m and 4 divides n(trivial since we can do 3x4-rectangles), 12 divides m, then n can be arbritarily, but not equal to 1,2 or 5(it is clear that n is not 1,2, and 5 is easy to exclude. in all other cases we may write n as linear combination of 3 and 4 with positive coefficients and we can use 3x4-rectangles again taking rows of 3x4-rectangles or 4x3-rectangles). so we are left with the case 6 | m and 2 | n. so look again at our involution: it shows that our hooks appear in pairs of the possible forms: either 3x4-rectangles or something similar to an S: .=== .=== ===. ===. now mark every second row. it is easy to see that there are always 6 out of the 12 squares of the S are marked, the same for horizontal 3x4-rectangles, but vertical 3x4-rectangles have either 4 or 8 squares marked, that shows that there's an even number of them. doing the same with columns, we get that the total number of 3x4-rectangles is even. now, look at the coloring of .==. =..= .==. =..= and periodic. it is easy to see that 6 out of 12 squares are marked at 3x4-rectangles, the S upright, the S turned by 90 degree and even distance to the left boundary of the rectangle, but 4 or 8 for an S turned by 90 degree with odd distance to the left boundary. this shows that there's an even number of them and by translating the coloring by 1 to the right and turning it all around, we get that the number of S is even as well. this shows that 24 is a divisor of mn, so we get one of the two other cases. i only found all solutions but couldn't prove that it are all. Peter

Hi all, I am the author of this problem (or the one who proposed it actually, the solution is a teamwork). I thought that it might be interesting for the readers of this forum to know the story behind its birth. So here it comes. In autumn 2003 I gave a problem creation seminar for our young olympiad organizers, since they complained that they were not able to come up with new problems. As the first topic I chose combinatorics and said: "Problem creation in combinatorics is trivial -- take an arbitrary shape (I drew the hook above) and ask whether it is possible to fill some rectangle with these". I never imagined that this would ever become a real olympiad problem, yet to talk about IMO! Jan

Dear Jan, it would be great if you inform us where do you live, and to post the "official" solution to your problem. Thanks in advance. Kantor

I agree with Kantor: the more solutions the merrier. So please, Jan, post the solution(s) you had in mind. ~Mike

http://www.math.ucf.edu/~reid/Polyomino/j6_rect.html gives one proof. This is indeed an easy sort of problem to create if you don't mind the solution being already known and online.

"There is a small hole in the proof of Part 1, pertaining to Figure 3C. The astute reader should be able to find and close the hole." Edit: Actually, the hole was fatal. The proof has been removed. Sorry.

Here's a cute way of showing that if a rectangle $6a\times 2b$ can be tiled with $3\times 4$ rectangles then at least one of $a,b$ must be even: we start from a corner of the board and place a $1$ there. Then we continue to the right, placing $i\cdot k$ to the right of $k$ (so the numbers in the squares will be $1,i,-1,-i$). The $n+1$'th row is the $n$'th row multiplied by $i$ (so we write $i\cdot k$ below $k$). Each $3\times 4$ rectangle has the same number of occurences for $1,i,-1,-i$, but it's easy to see that a rectangle of the form $4m+2\times 4n+2$ doesn't satisfy this.

Well, the problem in itself is nice, but all these tiling-type problems have been studied a lot and many papers appeared on this kind of problems. Thus, the reference given by jsm28 is not so surprising and this probably appear somewhere else. This should be sufficient for the jury to eliminate the problem, because some contestants may have known it before. Pierre.

By some simple observations (but not trivial, in fact i think that the most important part is that of the invultion, there is the magic) the problem reduces to tile a rectangle of shape m*n (with m=6m and n=2nmand nodd) with 3*4 bricks and some S (which shape you probably know). What we want is to show that the number of S is even, and that the number of 3*4 bricks is even two. From here we deduce that 24 divide mn which contradict the shape of the rectangle. For the bricks, procede in this way, put the number i-j on the intersection of column i and row j. You can easy check that mod 12 the sum of every S is Zero and that of every brick is 6. But if you look every 2*6 rectangle have sum Zero mod 12, from where the whole rectangle have sum zero, so the number of bricks is even. For the S, we do the following, call a j-column a column numebr k with j=k mod 4. Put in each 1-column a 1 and in each 3-column a -1. It is easy to check that the sum of all number is 2 mod 4.But bricks have sum Zero mod 4, and the 3333 S too.Now, the S with 2442 square per column have sum +2 or -2 mod 4. Hence Suming over all S we get that there is an odd number of this type of S. For the other type of S (the one with 3333 squares per column) aply the same idea but with rows and we get that of this new type there are an odd number. Then the total number of S is even and we are done. Ok, when i wrote it i havent read yet that link solution, it make me feel like a fool such nice solution, anyway...here is what i did.

Hi all, Sorry for not specifying my country in the first post -- I (and hence also the problem) come from Estonia. I was afraid that something like the link posted by jsm28 might come up, but there is really no way of telling it before it actually does. Our own solution went pretty much along the lines of this link, so there is not much point in retyping it here. There are several details (e.g. there is also a third way of covering the "center" of this hexamino, but that does not work), but they can be sorted out. Of course there is a philosophical question "some students might have seen this link". But please consider that the amount of material on Internet is huge and the chance that some student has found Michael Reid's page and memorized the facts about J6 hexamino, is pretty small. Anyway, Michael Reid's page has a lot of information about other tile shapes as well, so why not remember all of them? Jan

I like this problem, and I have the feeling a lot of solutions will come up . Here's another proof that a board $4m+2\times 4n+2$ can't be tiled with J-hexomino's: First put $1,i,-1,-1$ in the squares of the board like in my first post. In each $3\times 4$ rectangle the elements add up to $0$, so they have $M_4+M_4i$, while in each shape formed by two intertwined hexomino's such that they don't form a $3\times 4$ rectangle (call this shape $S$) the sum is $2,2i,-2$ or $-2i$. The sum of the numbers on the entire board is $2,2i,-2$ or $-2i$ (depending on the number we put in the top left corner), so there must be an odd number of $S$ shapes $(*)$. If the two $2\times 1$ bits of an $S$ shape are vertical we call that a vertical $S$ shape, and if they're horizontal we call it horizontal. We now put $i$ on the first line of the board, and continue by putting $i^n$ on the $n$'th line. It's easy to see that the sum of the numbers in the board has form $M_4+2+(M_4+2)i$, and the sum of the numbers in a $3\times 4$ rectangle and in a vertical $S$ is $M_4+(M_4)i$, while for a horizontal $S$ the sum is $M_4+2+(M_4+2)i$, so we must have an odd number of horizontal $S$. By doing the same, but with columns instead of lines, we find an odd number of vertical $S$, so the total number is odd+odd=even, contradicting $(*)$.

I didn't say that Jan. Your problem is nice, but I thought that an IMO problem had to be not 'well-known' (even if I didn't know it before myself). In my opinion, tiling problems as this one have given a lot of papers, with quite often elementary solutions, the problem is quite always to find the good invariant/coloration (which indeed can be really hard) and then it probably appeared in some math magazine somewhere, and thus may be known to some contestants. This risk is sufficiently high to think about it. Pierre.

I totally agree with Pierre, however the problem is very nice. Anyway, congratulations Jan for your second problem to be considered for an IMO, the first was problem 1 IMO 1999. Kantor

There was a major mistake in my last post, but it's been edited now.

And 1999/1, was really nice too. Congratulations Jan. Pierre.

grobby: Your proof doesn't work - You only showed that hooks appear in pairs on edges. Consider if the corner is tiled this way: 1112X 1212X 1222X 333PX 443XX 433XX 444XX XXXXX ..... (different numbers represent different hooks.) And your proof doesn't say what would happen at P. With more hooks added this can go really complex. Here's a brief of my proof: 1. Observe that "hooks" always appear in couples ie if a retangle is composed by hooks it can only be done in two ways: XXXXXX XX111X XX121X X212XX X222XX XXXXXX or XXXXX X1112X X1212X X1222X XXXXX where 1, 2 are two hooks. This shows that mn must be divisble by 12. 2. 3m*4n is always fine since it's composed of 3*4 rectangles which is composed of two hooks. 3. 12m*5 is impossible. 4. 12m*7 is possible by dividing it into 4m pieces of 3*4 rectangles and 3m pieces of 4*3 rectangles. 5. 12m*6, 12m*8 is possible by (2). Then 12m*n for all n>5 is possible by dividing it into a 12m*(n-3) rectangle and a 12m*3 rectangle. 6. 6m*2n is not possible for odd m, n, by coloring the small squares of the rectangle with four colors according to parity of coordinates: (odd, odd): color 1 (odd, even): color 2 (even, odd): color 3 (even, even): color 4 Then each "hook couple" in (1) always covers an even number of squares of each color, but a 6m*2n rectangle has 3m*n (odd) squares of each color

Dude, I don't think I understand what you mean.. I used $3\times 4$ rectangles and $S$ shapes, which correspond to your second and respectively first picture. I think the solution shows that a rectangle tiled with $3\times 4$ rectangles and $S$ shapes can't have dimensions $4m+2\times 4n+2$. What's wrong? I didn't try to show that hooks must form $3\times 4$ rectangles..

grobber: Maybe I misunderstood, but you said "First, I prove that the hooks must be combined to form 3 x 4 rectangles." in the first line

The only thing I do not get about this question is: why did they make it number 3?? I mean I've already tried day 1 at home, and I guess I'd have gotten a 7 for this one (which I'd certainly not get for the others). It's simply 1. showing they can only come as rectangles or semi-tetros (the other form), this means 12 | mn (pretty easy) 2. showing it's ok if at least 1 side is a multiple of 4 (even easier) (unless you got a 12x1 or 12x2 or 12x5 case -- the other number must be writable as 4m+3n) 3. finding any contradiction on the way that no side divides 4. (I did the the same way as grobber: 1,i,-1,-i,1,... with sum and product contradicting.) Is it really an easy question to IMO standards, or am I just lucky having found the right coloring immediately? I'm not used to solve problems being number 3!

Valentin Vornicu wrote: Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure. Invalid image file Determine all $m\times n$ rectangles that can be covered without gaps and without overlaps with hooks such that - the rectangle is covered without gaps and without overlaps - no part of a hook covers area outside the rectagle. Can the squares move randomly or they hook is regular Abdurashid

Valentin Vornicu wrote: Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure. Invalid image file Determine all $m\times n$ rectangles that can be covered without gaps and without overlaps with hooks such that - the rectangle is covered without gaps and without overlaps - no part of a hook covers area outside the rectagle. Can the squares move randomly or the hook is regular Abdurashid

You have that hook, given, a lot of exact copies of it. So 6 blocks glued together. And by using only such hooks, you should overlap the m x n rectangle.

Can this problem be done using some idea of generating functions? That is, for the $ m\times n$ board, you consider each cell of the form $ x^i y^j$ where row wise you name the cell $ x^i$ and column wise you name the cells $ y^j$ and name the cells of the board correspondingly as $ x^iy^j$. For the hook, there are lots of cases leading to lots of construction of polynomial functions??? Can someone suggest an easier way.

Sorry, I don't understand what Grobber means by Msub4. Could someone please explain?

grobber wrote: I like this problem, and I have the feeling a lot of solutions will come up . Here's another proof that a board $4m+2\times 4n+2$ can't be tiled with J-hexomino's: First put $1,i,-1,-1$ in the squares of the board like in my first post. In each $3\times 4$ rectangle the elements add up to $0$, so they have $M_4+M_4i$, while in each shape formed by two intertwined hexomino's such that they don't form a $3\times 4$ rectangle (call this shape $S$) the sum is $2,2i,-2$ or $-2i$. The sum of the numbers on the entire board is $2,2i,-2$ or $-2i$ (depending on the number we put in the top left corner), so there must be an odd number of $S$ shapes $(*)$. If the two $2\times 1$ bits of an $S$ shape are vertical we call that a vertical $S$ shape, and if they're horizontal we call it horizontal. We now put $i$ on the first line of the board, and continue by putting $i^n$ on the $n$'th line. It's easy to see that the sum of the numbers in the board has form $M_4+2+(M_4+2)i$, and the sum of the numbers in a $3\times 4$ rectangle and in a vertical $S$ is $M_4+(M_4)i$, while for a horizontal $S$ the sum is $M_4+2+(M_4+2)i$, so we must have an odd number of horizontal $S$. By doing the same, but with columns instead of lines, we find an odd number of vertical $S$, so the total number is odd+odd=even, contradicting $(*)$. what do you mean by M sub4?

riemann, I'm pretty sure that this famous Msub4 means just "multiple of $4$", with classical notation $\mathcal{M}4$ (grobber talks about "... form $M_4$ ...). But do you noticed you are asking a question on a 2004 or 2005 post? Never too late ... !

The answer is: $4a \times 3b$ rectangles (with $a,b \ge 1$) and $12a \times b$ rectangles (with $b \neq 1,2,5$ and $a \ge 1$). Proof: Note that we can unite two hooks to form a $3\times 4$ rectangle. Using these rectangles it's easy to see $4a \times 3b$ rectangles (with $a,b \ge 1$) satisfy. Now, if we have a $12a \times b$ rectangle (with $b \neq 1,2,5$ and $a \ge 1$), then by stacking either horizontally or vertically the rectangles, we see that the rectangle can be tiled if a $12a \times (b-3)$ or $12a \times (b-4)$ rectangle can be tiled. So if $b=4x+3y$ ($x,y \ge 0$), then the rectangle can be tiled. It's easy to see that all numbers $b$ are of that form, except for $b=1,2,5$. Therefore, the rectangles I described above can be tiled. Now I prove no other rectangle can be tiled. Assume $n \times m$ is not one of the above rectangles, and that it can be tiled (we will prove a contradiction). Note that the hole of each hook must be covered by some other hook, and it's easy to see we can pair up the hooks like this. This forms $3\times 4$ rectangles and strange "snape" shapes, which look like two 2x3 rectangles stacked unevenly. Both of these shapes have 12 squares, and so $12 | nm$. Since $n,m$ is not of the above rectangles, then we have that $n \times m$ = $12k \times 1$, $12k \times 2$, $12k \times 5$ or $6a \times 2b$, with $a,b$ odd and $k$ a positive integer. In the last case ($6a \times 2b$), color the rectangle like a chess board, with the top left corner black. We have eight types of snapes: $S_1$ snapes: oXxx oxxx xxxo xxXo where the capital $X$'s are black. $S_2$ snapes: xxxo xxXo oXxx oxxx where the capital $X$'s are black. $S_3$ snapes: ooxx Xxxx xxxX xxoo where the capital $X$'s are black. $S_4$ snapes: xxoo xxXx xXxx ooxx where the capital $X$'s are black. $S_5$ snapes: oXxx oxxx xxxo xxXo where the capital $X$'s are white. $S_6$ snapes: xxxo xxXo oXxx oxxx where the capital $X$'s are white. $S_7$ snapes: ooxx Xxxx xxxX xxoo where the capital $X$'s are white. $S_8$ snapes: xxoo xxXx xXxx ooxx where the capital $X$'s are white. Now, color the rectangle with $4$ colors, using southwest-northeast diagonals. The first diagonal (which in reality is only the square in the top left corner) is color $1$, the second diagonal is color $2$, third is color $3$, fourth is color $4$, fifth is color $1$, and so on. We see that there will be $Y$ squares $1$, $Y+1$ squares $2$, $Y$ squares $3$ and $Y-1$ squares $4$, for some $Y$. Each $3 \times 4$ rectangle will cover 3 squares of each color. For all $k \ge 5$, each $S_k$ snape will cover 3 squares $1$ and 3 squares $3$. For all $k \le 4$, each $S_k$ snape will cover 2 squares $1$ and 4 squares $3$, or 4 squares $1$ and 2 squares $3$. Since the number of squares $1$ is equal to the number of squares $3$, we have $S_1+S_2+S_3+S_4$ is even. Now, color again, but this time using southeast-northwest diagonals, so that the top left corner is color $1$. This time, the number of squares $2$ is equal to the number of squares $4$. With analogous reasoning, we see that $S_5+S_6+S_7+S_8$ is even. Therefore, the number of snapes is even. Finally, color with 4 colors, according to the parity of each coordinate of the squares. Each snape will cover 3 squares of each color, while each $3 \times 4$ rectangle will cover an even number of squares of each color. Since in total there are an odd number of squares of each color, there will be an odd number of snapes! Contradiction. Therefore we have $n \times m$ = $12k \times 1$, $12k \times 2$ or $12k \times 5$ Since each hook takes up space from 3 rows and 3 columns, the first two options are not possible. Now if $n=12k$, $m=5$, with $k$ a positive integer, look at the first column. (Columns have 5 squares). We consider each $3 \times 4$ rectangle and each snape as the union of two $2 \times 3$ rectangles. The top left and bottom left corners can't both be covered by vertical rectangles. If they are both covered by horizontal rectangles, then the 3rd square of the left-most column can't be covered. So, without loss of generality the top left corner is covered by a horizontal rectangle, and the bottom left is covered by a vertical one. Then it's easy to see that that horizontal rectangle can't be united with any other rectangle. Contradiction! So we are done. The answer is: $4a \times 3b$ rectangles (with $a,b \ge 1$) and $12a \times b$ rectangles (with $b \neq 1,2,5$ and $a \ge 1$).

How about coloring the board in 12 colors ABCDABCD... EFGHEFGH... IJKLIJKL... on below and sideways. I think the rest is clear of as each square covers only 1 of the colors.

Edit: this is a fakesolve

For the $6a \times 2b$ case, I have a short coloring proof which I haven't seen anywhere above. I provide a complete solution: By the manner in which the "hole" can be filled, there are two possible shapes: an S-shape and a $3 \times 4$ rectangle. These are the only two ways to cover the hole, and the holes of both participating hooks are filled, so hooks occur in pairs and are even in number, so $12 | mn$. Clearly $3|m$ and $4|n$ works, since this can be tiled with $3 \times 4$ rectangles. $12|m$ and $n \neq 1,2,5$ also works since $12 \times 3$ and $12 \times 4$ are possible. These can be placed one above the other to give $n$ which is a linear combination of $3$ and $4$, clearly giving $n \neq 1,2,5$, and all others work. The $12 \times 5$ rectangle can be shown to be un-tilable using some small casework for corner pieces. There remains one case: $6a \times 2b$ for odd $a, b$. Note that $4$ doesn't divide either of $m, n$. Assume it is possible to fill the grid with hooks. Colour the grid with colours $1$ through $4$ with colour congruent to sum of row and column number modulo $4$. In colouring a $6a \times 2b$ rectangle for odd $a, b$, there is $1$ more square with colour $1$, $1$ more square with colour $3$, and $2$ more squares with colour $2$ than squares of colour $4$. However, in every S-shape and $3 \times 4$ rectangle, the parity of number of $2$-coloured and $1$-coloured squares is always the same, hence their difference cannot be $1$ when summed over several S-shapes and $3 \times 4$ rectangles, a contradiction.

Clearly, if there exists a hook, then the square in the middle must be part of another hook. The only valid options are the following two: [asy][asy] size(6cm,0); draw((0,0)--(3,0)--(3,3)--(2,3)--(2,1)--(1,1)--(1,2)--(0,2)--(0,0)); draw((1,2)--(1,4)--(4,4)--(4,2)--(3,2)); draw((6,0)--(6,3)--(10,3)--(10,0)--(6,0)); draw((7,0)--(7,2)--(8,2)--(8,1)--(9,1)--(9,3)); [/asy][/asy] In particular, there must be an even number of hooks, so the number of grid is a multiple of $12$. From now on, we will ban rotations and instead add two more figures that we can tile with. Reflections are still allowed. Denote the first one by $A$ and its ninety degree rotation by $B$. Denote the second one by $C$ and its ninety degree rotation by $D$. If we color every other row red, each type of figure will cover six red squares, except figure $B$, which will cover four or eight. Similarly, if we color every other column red, each type of figure will cover six red squres, except figure $D$, which will cover four or eight. If we group two squares horizontally and apply chessboard to those groups, each type of figure will cover six red squares, except figure $C$, which will cover four or eight. If we do the same vertically we get similar result for $A$. Thus, if both $m$ and $n$ are even, then we have proved that an even number of each figure is needed, so $24\mid mn$. Then, one of them must be divisible by $4$. If not both $m$ and $n$ are even, $12\mid mn$ forces that anyway. WLOG, assume $4\mid m$. Then, either we have $12\mid m$ or $4\mid m$, $3\mid n$. The second one is obviously possible and the first is possible for any $n=3a+4b$ for some nonnegative integers $a,b$. That is, $n$ can equal anything that isn't $1$, $2$, or $5$ and it's each to check that those don't work.

The answer is all $3a \times 4b$ rectangles, as well as all $12a\times b$ rectangles, where $b \not \in \{1,2,5\}$. To see that these work, observe that we can use $3 \times 4$ rectangles only to tile the first class, and for the second we can use $3 \times 4$ rectangles to form both $3 \times 12$ and $4 \times 12$ rectangles, so by Chicken McNugget we can form any $12a \times b$ rectangle for $b \geq 6$. It is also easy to see that we can form $12a \times 3$ and $12a \times 4$, but clearly $b=1,2,5$ are not possible. We now show that these are the only answers. It is clear that in any valid tiling, by considering the "divot" in each hook, we can either pair up the shapes to form a $3 \times 4$ rectangle or an "S-shape", which is a $4 \times 4$ square with the highest two cells on the left column and the lowest two cells on the right column removed, as well as rotations and reflections. This immediately implies that $12 \mid mn$. The rest of this solution will show that we cannot have $\nu_2(m)=\nu_2(n)=1$; suppose that this was true for the sake of contradiction. Ban rotations (but allow reflections) and consider the four following shapes: Rectangles which are $3$ wide and $4$ tall (type A) Rectangles which are $4$ wide and $3$ tall (type B) S-shapes where we remove from the left/right columns (type C) S-shapes where we remove from the top/bottom rows (type D) Number the columns $0$ through $n-1$ in order, and for $0 \leq i \leq 3$ let $a_i$ denote the number of type A rectangles whose leftmost cells lie in a column whose label is congruent to $i \pmod{4}$. Define $c_i$ similarly. Now, By coloring the columns in the pattern $1212\ldots$, we get an equal number of $1$s and $2$s. Type B, C, and D shapes cover an equal number of $1$s and $2$s, so we get the equation $a_0+a_2=a_1+a_3$. By coloring the columns in the pattern $1221\ldots$, we get an equal number of $1$s and $2$s, since $\nu_2(n)=1$. Type B and D shapes cover an equal number of $1$s and $2$s, but type A shapes either have four extra $1$s or four extra $2$s, and certain C shapes can as well, so we get the equation $a_0+a_1+c_0=a_2+a_3+c_2$. By coloring the columns in the pattern $1122\ldots$, we get $2m$ extra $1$s; note that $\nu_2(2m)=2$. Type B and D shapes still cover an equal number of $1$s and $2$s, but type A shapes either have four extra $1$s or four extra $2$s, and certain C shapes can as well, so we get the equation $a_0+a_3+c_3=a_1+a_2+c_1+\tfrac{m}{2}$. Taking things modulo $2$, it follows that $c_0 \equiv c_2 \pmod{2}$ and $c_3 \equiv c_1+1 \pmod{2}$, since $\tfrac{m}{2}$ is odd. Therefore there are an odd number of type C shapes. Also, since $a_0+a_1+a_2+a_3 \equiv 0 \pmod{2}$, there are an even number of type A shapes. By repeating this argument with rows and type B/D shapes, we also get that there are an odd number of type D shapes, and an even number of type B shapes. Therefore, there are an even total number of shapes, so the number of cells in the $m \times n$ rectangle must be divisible by $24$, but this contradicts $\nu_2(m)=\nu_2(n)=1$; done. $\blacksquare$ Remark: As a wise man once said, Inconsistent wrote: This problem is similar to modern art, in the sense that I have no clue what the point of this problem even is.

I think my sol is one of the contenders for worst coloring award. Note that any such configuration must have $12 \mid mn$ because there's only two ways to put another piece into the hook of another hook: either as a $3 \times 4, 4 \times 3$, or a $4 \times 4$ with two dominos removed from either corner of the same orientation. As such, it follows that $12 \mid mn$. Constructions Claim: This is possible for $3a \times 4b$. Proof. Tile with $3 \times 4$s. $\blacksquare$ Claim: This is possible for $12a \times b$ iff $b \not\in \{1, 2, 5\}$. Proof. $b \in \{1, 2, 5\}$ is easily checked as impossible. For other $b$, we can decompose $b = 4a_1 + 3a_2$ for nonnegative integers $a_1, a_2$, which can be tiled as before. $\blacksquare$ Impossibility We now show it's impossible for $6a \times 2b$ where $a, b$ are odd. Call a piece which consists of a $4 \times 4$ without corners a silly piece. We color the piece based off $(x \pmod{4}, y \pmod{4})$ such that that the bottom left corner is the following square. [asy][asy] size(5cm); int N = 4; for (int i = 0; i <= N; ++i) { draw((0,i)--(N,i)); draw((i,0)--(i,N)); } pen r = red+white+white; fill(shift(0,0)*unitsquare, r); fill(shift(1,0)*unitsquare, r); fill(shift(1,1)*unitsquare, r); fill(shift(2,1)*unitsquare, r); fill(shift(2,2)*unitsquare, r); fill(shift(3,2)*unitsquare, r); fill(shift(3,3)*unitsquare, r); fill(shift(0,3)*unitsquare, r); [/asy][/asy] Then each $3 \times 4, 4 \times 3$ covers an even number of red squares and each silly piece covers an odd number of red squares. Since there are in total an odd number of red squares, there are an odd number of silly pieces. On the other hand, consider the following coloring. [asy][asy] size(5cm); int N = 4; for (int i = 0; i <= N; ++i) { draw((0,i)--(N,i)); draw((i,0)--(i,N)); } pen r = red+white+white; fill(shift(0,0)*unitsquare, r); fill(shift(3,0)*unitsquare, r); fill(shift(1,1)*unitsquare, r); fill(shift(2,1)*unitsquare, r); [/asy][/asy] Every piece covers an odd number of red squares except for silly pieces which have dominos removed horizontally and bounding box's leftmost horizontal residue is $1 \pmod{2}$, call this set of pieces $V_1$ with its subscript as its residue and $V$ for vertical. Define $V_1, H_0, H_1$ similarly. Since there are in total $ab \equiv 1 \pmod{2}$ pieces and the coloring places an even number of red squares (consider each $2 \times 2$ separately), this implies that there are an odd number of pieces in $V_1$. By similar logic with a translated / reflected coloring, we get $V_0, H_0, H_1$ have an odd number of pieces as well. However, the total number of silly pieces is the sum of $V_0, V_1, H_0, H_1$, which is odd, contradiction.