Suppose it is possible to tile the board with $16$ strominos such that each square is covered at most twice. We count for each square the number of times it is covered, suppose $n$ squares are covered only once; then $n+2 \cdot (25-n)=50-n$ is the sum of all these counts. On other hand, each stromino contributes $3$ to this sum, so $50-n=3 \cdot 16$ so $n=2$. Label the rows and columns from $1$ to $5$. Let $\omega=e^{\frac{2\pi i}{3}}$ be a cube root of unity. Assign to square $(i, j)$ the value $\omega^{i+j}$. Consider the sum over all squares where each square contributes its value times the number of occurrences. This totals $2\left(\sum_{k=1}^5 \omega^k\right)^2-\omega^{x+y}-\omega^{z+t}$ where $(x, y)$ and $(z, t)$ are the cells not covered twice. However, some of values for each stromino is zero, hence this sum should be zero. So $\omega^{x+y}+\omega^{z+t}=2 \implies x+y \equiv z+t \equiv 0 \pmod{3}$. Now again, do the same trick while colouring $(i, j)$ by $\omega^{i-j}$. We conclude that $\omega^{x-y}+\omega^{z-t}=2 \implies x-y \equiv z-t \equiv 0 \pmod{3}$. Thus, $3$ divides each element of $\{x, y, z, t\}$ and as they all lie in $[1, 5]$ we get that $x=y=z=t=3$. But then, the square $(3,3)$ is the only square not covered twice; contradicting that there are two of them. $\blacksquare$

anantmudgal09 wrote: Suppose it is possible to tile the board with $16$ strominos such that each square is covered at most twice. We count for each square the number of times it is covered, suppose $n$ squares are covered only once; then $n+2 \cdot (25-n)=50-n$ is the sum of all these counts. On other hand, each stromino contributes $3$ to this sum, so $50-n=3 \cdot 16$ so $n=2$. Label the rows and columns from $1$ to $5$. Let $\omega=e^{\frac{2\pi i}{3}}$ be a cube root of unity. Assign to square $(i, j)$ the value $\omega^{i+j}$. Consider the sum over all squares where each square contributes its value times the number of occurrences. This totals $2\left(\sum_{k=1}^5 \omega^k\right)^2-\omega^{x+y}-\omega^{z+t}$ where $(x, y)$ and $(z, t)$ are the cells not covered twice. However, some of values for each stromino is zero, hence this sum should be zero. So $\omega^{x+y}+\omega^{z+t}=2 \implies x+y \equiv z+t \equiv 0 \pmod{3}$. Now again, do the same trick while colouring $(i, j)$ by $\omega^{i-j}$. We conclude that $\omega^{x-y}+\omega^{z-t}=2 \implies x-y \equiv z-t \equiv 0 \pmod{3}$. Thus, $3$ divides each element of $\{x, y, z, t\}$ and as they all lie in $[1, 5]$ we get that $x=y=z=t=3$. But then, the square $(3,3)$ is the only square not covered twice; contradicting that there are two of them. $\blacksquare$ Exactly my solution, but for the second labelling I have used, $\omega^{x+2y}$. Though both are same.

Due to similar reasoning, 2 squares would be left covered once and rest twice. Colour the board as follows 12312 23123 31231 12312 23123 Now there are 9 2's and 8 1's and 8 3's. So in colouring, all 1's, 2's and 3's would be covered twice but 2 2's would be left. Now consider colouring 54654 65465 46546 54654 65465 Here also due to similar reasoning, 2 5's will remain. For a specific covering, if we cover both the boards with same cover, then same squares would be left, that would be the intersection of 5's and 2's. But it is a unique square (centre), contradicting 2 squares are left.

Like these problems are abundNt in pftb!!

I had the same start as #2 , the first two paragraphs exactly match! But then I had to silly it.

I showed that there can’t be $ \geq 10$ horizontal/vertical strominos. Then I showed that 9 of one type/ 7 of other wouldn’t hold. Then I showed that 8/8 doesn’t hold either. Basically disproved the only possible cases Did anybody do the same?

Here is what l did Label the board like 12345 23451 34512 45123 51234 Every stromino contains exactly 3 different numbers Let $S_1$ denote the number of total strominos on number 1 and similar definition for $S_2$ and so on We observe that any stromino contains at least 2 consecutive numbers (look at the labelling). Hence 135 stromino does not appear anywhere. Note that this also means that after we place a stromino in any any three unit squares the quantity L=$S_1+S_3+S_5-S_2-S_4$ changes by one with every tile( since 2 numbers are consecutive in them 1 odd and 1 even). We consider all possible $S_1,S_2,S_3,S_4,S_5$ distribution. Each $S_i$ <11 otherwise there is by php one among the label i that is covered thrice. $S_1,S_2,S_3,S_4,S_5$ ={10 10 10 9 9} Or {10 10 10 10 8} in some permutation. One useful observation is that in any case there are t least 8 times that a number is covered. Which means at most twice not covered We calculate the quantity E which counts the number of 234 covering stromino. Let E be (16-L)/2 since L represent by how much the sum of odd Si i.e $S_1,S_3,S_5$ is greater than even $S_i$, i.e $S_2+S_4$. Note that a tile increase even S2+S4 by one compared to $S_1+S_3+S_5$ only if 234 is the stromino and it increases $S_1+S_3+S_5$ by one compared to $S_2+S_4$ only is it is (odd,odd,even) . Then 16-L reprensents the number of times that (odd,even,even)covering stromino in any order is equal to (odd even odd) in any order . So dividing by 2 gives us the (even even odd) case which only happens to be 234 and moreover every 3 contains a 234 string . Hence $E\le2$ for any combination otherwise we can have at most 7 times number 3 is used. But $E\ge2$ for each combination with equality when {10, 10,10,8,10} or {10,9,10,9,10} in either case we already assume $S_3=10$ contradiction. Plz someone tell if it is completely correctly. I would be thankful. But I messed writing this clearly. I home examiner understand it

SRIRAMACHANDRA wrote: Like these problems are abundNt in pftb!! What is PFTB

PDT2020 wrote: SRIRAMACHANDRA wrote: Like these problems are abundNt in pftb!! What is PFTB Problems From The Book, By Titu Andreescu

RudraRockstar wrote: I showed that there can’t be $ \geq 10$ horizontal/vertical strominos. Then I showed that 9 of one type/ 7 of other wouldn’t hold. Then I showed that 8/8 doesn’t hold either. Basically disproved the only possible cases Did anybody do the same? Meeeeee with 5 page long solution.

RANDOMMATHLOVER wrote: RudraRockstar wrote: I showed that there can’t be $ \geq 10$ horizontal/vertical strominos. Then I showed that 9 of one type/ 7 of other wouldn’t hold. Then I showed that 8/8 doesn’t hold either. Basically disproved the only possible cases Did anybody do the same? Meeeeee with 5 page long solution. Ok you just prevented me from going to the Himalayas :relieved:

Call a stromino vertical if placed vertically, and horizontal if horizontally. Note that any horizontal stromino passes through the center of its row, and any vertical stromino through its corresponding column. Call a square "blocked" if 2 strominoes already pass through it, aka we can't put another stromino covering that square. Clearly, there can be at most 2 vertical strominoes in the same column, and 2 horizontal strominoes in same row, thus there are at most 10 vertical and 10 horizontal strominoes. Assume to the contrary. Now, without loss of generality, let number of vertical strominoes $\geq$ number of horizontal strominoes. Note that at least 3 of the column centers are blocked, as there are $\geq 8$ number of vertical strominoes, at most 2 vertical strominoes in a column, and 5 columns. Thus we get that there must be no horizontal stromino in middle row (as 3 of middle rows' squares are blocked). Here we take cases: Case 1: There is at least one vertical stromino in middle column. Say it covers 2 rows, $R_i$ and $R_j$ except for middle row. Clearly there can be at most 1 horizontal stromino in each of the rows $R_i$ and $R_j$, hence at most $2 \cdot 2+2 \cdot 1=6$ horizontal strominoes. Thus there must be 10 vertical strominoes and 6 horizontal strominoes- but this implies 2 vertical strominoes in middle column. But as we are adding another stromino in middle column, thus we're reducing the number of possible horizontal strominoes by 2 $\implies$ at most 4 horizontal strominoes (I actually case-bashed this in the exam instead of simply using maximum number of possible horizontal strominoes $- = 2$). Thus contradiction. Case 2: No vertical stromino in middle column. Contradiction to occupancy of every square, as we proved no stromino in middle column, and no vertical stromino in middle column, hence done.

Solution: Assume to the contrary that such a configuration is possible. Now as total number of tiles in strominoes is $3\times 16=48$ and each of the $25$ tiles are part of $1$ or $2$ stromino, we get that exactly $2$ tiles must be part of only $1$ stromino. We reach our desired configuration through a set of moves, where each move consists of adding a stromino to the board starting from an empty board. Now fill the board as 1 0 2 1 0 0 2 1 0 2 2 1 0 2 1 1 0 2 1 0 0 2 1 0 2 First fix $A=0$ Now on on each move add the corresponding numbers under our stromino to $A$ . Then on each move $A$ increments by $3$, so for the desired configuration $A=16\times 3=48$ If we add all the numbers twice we get $48$ , so the two numbers corresponding to the tiles under only $1$ stromino will be $0$. Hence those two tiles will be in place of $0$ only. Now using same logic for the mirror image of the numbering we get that those two tiles are again in place of $0$ only. But the intersection of the places of $0$ in both numbering is only one. Contradiction. Hence proved. @2 below, Amazing solution!!!

Why does no one see that some INMO problem is forced to be seen in another contest before? See STEMS 2019 objective paper. Precisely the same solution as in #4 was used to solve it. Needless to say, it works here too. @below: Nice, that's cute

Since there are 16 strominos to be used, total squares that might be covered once or twice in the board is 48. Let n be the number of squares covered twice using strominos. We get 2(n) + 1(25-n) = 48. Solving we get, n = 23. Define a DOUBLE STROMINO to be any arrangement of a board that can be covered using just strominos and each unit square is covered by two strominos. CLAIM: A double stromino only exists if the total number of unit squares on the board is a multiple of 3. PROOF OF THE CLAIM: let the total number of unit squares be k. The total squares used by strominos will be 2k. Since 3 does not divide 2. 3 must divide k. We see that for our problem we need a double stromino for 23 unit squares. By our claim 23 should be a multiple of 3. Since it is not, any arrangement of double stromino for 23 will not exist. Hence we get the desired result that the 5x5 board cannot be covered using 16 strominos

Nothing too different from the previous solutions, but I am posting it anyway: Suppose $n$ squares are covered by exactly one stromino. Then, $$n + 2(25-n) = 3 \cdot 16$$which means $n = 2$. Label the squares from $(1,1)$ to $(5,5)$ in the obvious way. Colour the square $(i,j)$ with the number $(i+j) \pmod 3$. There are $9$ squares coloured $0 \pmod 3$, $8$ squares coloured $1 \pmod 3$, and $8$ squares coloured $2 \pmod 3$. Now, note that each stromino covers exactly one square of each colour. This means the two squares covered exactly once must be the ones coloured $0 \pmod 3$. Now, take a second colouring, where the square $(i,j)$ is coloured $i-j \pmod 3$. By a similar logic as before, the two squares covered exactly once must be the ones coloured $0 \pmod 3$. However, the only square $(i,j)$ with $i+j \equiv 0 \equiv i-j \pmod 3$ is $(3,3)$, which is a contradiction. Therefore, a tiling with the required conditions is impossible.

MathPassionForever wrote: Why does no one see that some INMO problem is forced to be seen in another contest before? See STEMS 2019 objective paper. Precisely the same solution as in #4 was used to solve it. Needless to say, it works here too. @below: Nice, that's cute Please mention the category and the question number.

Pathological wrote:

This is an amazing solution.

$\mathrm {WLOG} $ assume that it is possible to do so that we number the squares in the grid as follows $\boxed {0}\boxed {1}\boxed {2}\boxed {0}\boxed {1}$ $\boxed {1}\boxed {2}\boxed {0}\boxed {1}\boxed {2}$ $\boxed {2}\boxed {0}\boxed {1}\boxed {2}\boxed {0}$ $\boxed {0}\boxed {1}\boxed {2}\boxed {0}\boxed {1}$ $\boxed {1}\boxed {2}\boxed {0}\boxed {1}\boxed{2}$ There are $9$ squares are numbered $1$, $8$ squares are $0$ and $8$ squares are numbered $2$. Each grid $(3\times 1) $ or $(1\times 3)$ covers exactly one square numbered $0$, one square is numbered one and one square is numbered $2\implies $ All the squares are numbered $1$ must be covered by $2$ grids $(3\times 1) $ or $(1\times 3)$. Consider a different grid of numbering $\boxed{0}\boxed {2}\boxed {1}\boxed {0}\boxed {2} $ $\boxed {1}\boxed {0}\boxed {2}\boxed {1}\boxed {0} $ $\boxed {2}\boxed {1}\boxed {0}\boxed {2}\boxed {1} $ $\boxed {0}\boxed {2}\boxed {1}\boxed {0}\boxed {2} $ $\boxed {1}\boxed {0}\boxed {2}\boxed {1}\boxed {0} $ Similar as above, it is easy to see that only squares numbered $0$ maybe covered by one grid $(1\times 3) $ or $(3\times 1)$. All squares in the grid($\textbf {except from the square in the center} $) must be covered by $2$ grids $(1\times 3) $ or $(3\times 1)$. The other squares must be covered twice. We use $48$ squares. Hence, the assumption is incorrect, and we can't use $16$ strominoes such that every square is covered by $1$ or $2$ strominoes.

Whenever a unit square gets a contribution from a stromino, we add $1$ to the square (we start with $0$ on all squares). It follows that in the end, we will have $1$s and $2$s in all squares. Also, if there are $k$ ones and $25-k$ twos, then $k+2(25-k)=3\times 16=48$, so $k=2$. Also, let $V$ denote the number of vertical strominos, and $H$ denote the number of horizontal strominos. Now consider the middle row, middle column, and the middlemost square. We note the following: $\bullet$ The number on the middlemost square indicates the number of strominos placed either vertically on the middle column, or horizontally on the middle row. Hence, there are at least one and at most two such strominos. $\bullet$ The sum of the numbers on the middle row minus $S$, where $S$ is thrice the number of strominos placed horizontally on the middle row equals $V$ (since all vertical strominos will touch the middle row). The value of $S$ can be either $0,3$ or $6$, due to the previous point. Same for the middle column (with the value being equal to $H$). The maximum value of $V$ (and $H$) can be $2+2+2+2+2=10$. We also know that $V+H=16$. WLOG, $V\geq H$. Then the possible values of $(V,H)$ are $(10,6)$, $(9,7)$ or $(8,8)$. If the middle row had at least one horizontal stromino in it, then the maximum value of $V$would have been $10-3=7$. Hence there are no horizontal strominos in the middle row. If $S_r$ and $S_c$ denote the sum on the middle row and column respectively, then $V=S_r$. Note that $\max(S_r)=\max(S_c)=10$. If the middlemost square has $1$, there is one vertical stromino on the middle column and the maximum value of $H$ is $10-1\times 3=7$. So $(8,8)$ is never possible. Now if the middlemost square has $2$, there are two vertical strominos on the middle column and the maximum value of $H$ is $10-2\times 3=4$. So $(10,6)$ is not possible, because if $V=S_r=10$, then the middle square is forced to have $2$. Now consider $(9,7)$. We must have $1$ on the middlemost square. But then the maximum value of $H$ will be $(2+2+2+2+1)-1\times 3=6$. So again, $(9,7)$ cannot occur. Hence, the tiling is not possible. $\blacksquare$

Can two storminoes be placed directly over each other?(coincident?)

At first of all let assigne a $5\times 5$ matrix with $a_{ij} = (i+j)$. Now notice that $\sum_{i=1} ^ 5 \sum_{j=1}^5 a_{ij} \equiv 0 \pmod{3}$. \begin{tabular}{lllll} \hline \multicolumn{1}{|l|}{{\color{blue} 2}} & \multicolumn{1}{l|}{\color{red} 3} & \multicolumn{1}{l|}{\color{green}4} & \multicolumn{1}{l|}{\color{black}5} & \multicolumn{1}{l|}{\color{blue}6} \\ \hline \multicolumn{1}{|l|}{{\color{red} 3}} & \multicolumn{1}{l|}{\color{green}4} & \multicolumn{1}{l|}{5} & \multicolumn{1}{l|}{ \color{blue}6} & \multicolumn{1}{l|}{7} \\ \hline \multicolumn{1}{|l|}{{\color{green}4}} & \multicolumn{1}{l|}{5} & \multicolumn{1}{l|}{ \color{blue}6} & \multicolumn{1}{l|}{7} & \multicolumn{1}{l|}{8} \\ \hline \multicolumn{1}{|l|}{{\color{black} 5}} & \multicolumn{1}{l|}{\color{blue}6} & \multicolumn{1}{l|}{7} & \multicolumn{1}{l|}{8} & \multicolumn{1}{l|}{9} \\ \hline \multicolumn{1}{|l|}{{\color{blue} 6}} & \multicolumn{1}{l|}{7} & \multicolumn{1}{l|}{8} & \multicolumn{1}{l|}{9} & \multicolumn{1}{l|}{10} \\ \hline & & & & \end{tabular}Now notice that , For each trominoes sum of the numbers that it cover is $0$ modulo $3$. Now let's assume that there are are$x$ unit square covered exactly once and $25-x$ unit square covered twice . So, $x+2(25-x) =16\times 3=48$. So there are only $2$ unit square coverd exactly once. Let assume that $a_{xy}$ and $a_{zw}$ are the unit squares that covered exactly once. Since, I have assume that the colouring is possible that's why , $2(\sum_{i=1} ^ 5 \sum_{j=1}^5 a_{ij})-a_{xy} -a_{zw} \equiv 0 \pmod{3}$. \[\implies (x+y)\equiv 2(z+w) \pmod{3} \cdots (1)\] again assign another kind of coloring , $a_{ij} = (i-j)$. In this case we also get, $2(\sum_{i=1} ^ 5 \sum_{j=1}^5 a_{ij})-a_{xy} -a_{zw} \equiv 0 \pmod{3}$. \[\implies x-y\equiv 2(z-w) \pmod{3}\cdots (2) \]. From equation (1 )and (2) we should get, \[x\equiv 2z \implies x+z \equiv 0 \pmod{3}\]. \[ \implies x+z \equiv y+w \equiv 0 \pmod{3}\]. Similarly we have $x-z \equiv y-w \equiv 0\pmod{3}$. So , only possibility is $x=y=z=w=3$. So there will exactly one unit square covered exactly once. Contradiction ! And we are done.

Perhaps a strong solution. Consider number of units squares covered by 1 stromino. It is given by (2*25-48)=2. The weakness lies in the dimensions of the square grid. The dimensions of square grid are 5 by 5 whereas the stromino's dimensions are 3 by 1. Hence, we get that every stromino covers at least one cell amongst the middle row or middle column (1).We number rows and columns as R_1, R_1... R_5 and C_1, C_2 ... C_5 in some order. Using (1), we we see that amongst 9 cells belonging to C_3 union R_3 = T, 16 strominos pass. At least one stromino passes through C_3 R_3 so that stromino in T, so 15 stromino pass through rest 8 cell (this is weaker case). See that amongst 8 cells, there are cell trough which only stromino can pass which makes that 13 strominos pass through rest 6 cells, a contradiction since 3 strominos cannot pass through a single cell but by PHP on the remaining 6 cells, at least one cell must be covered by 3 strominos. Suppose two strominos pass through C3R3. Then we have two cases, if stromino overlap then results 14 stromino pass through 6 cells, a contradiction and suppose they don't overlap,then 4 cell can be covered by one more stromino and so the last 4 stromino's are being covered by 10 stromino's, a safe contradiction. Aaaand we're done. Sorry my English and grammar is bad

anantmudgal09 wrote: A stromino is a $3 \times 1$ rectangle. Show that a $5 \times 5$ board divided into twenty-five $1 \times 1$ squares cannot be covered by $16$ strominos such that each stromino covers exactly three squares of the board, and every square is covered by one or two strominos. (A stromino can be placed either horizontally or vertically on the board.) Proposed by Navilarekallu Tejaswi Now let's number the row and column of the board starting from $1$ to $5$ respectively. Now observe middle square of any row or column appears in $3^\text{rd}$ column or row respectively. Middle Most row or column means $3^\text{rd}$ Row or column. Terminal row or column means $1^\text{st}$ or last row or column respectively. Now just observe as the square is $5\times 5$ So atmost $2$ Horizontal Strominos can cover each row By Satisfying Conditions and atmost $2$ Vertical Strominos can cover each Column Vertically. Claim 1-: We can't have More than $10$ Horizontal or Vertical Strominos. Prove-: FTSOC Assume that we can have more than $10$ Horizontal Strominos. Then there will be at least $1$ row which will have $>2$ Horizontal Strominos. Which is never possible as One square can only be covered atmost $2$ times. Similarly we can prove that we can't have $>10$ Vertical Strominos. Claim 2-: $16$ Strominos can never cover the Board, By satisfying given Conditions. Prove-: FTSOC Assume we can cover board by $16$ Strominos. Then we will have some possible cases. Now as Strominos are of $2$ types( Horizontal and Vertical). So there are some cases SubCase 1-: Trying to cover the board with $10$ of $1$ kind and $6$ Strominos of another kind. Prove-: Now to have $10$ Strominos of $1$ type(say Horizontal). We need to cover each row By exactly $2$ Strominos. And each there So there are $2$ Possibility to do so $1-:$ The Middle Most square of each Row can be covered twice and rest all once. $2-:$ Any $2$ Consecutive square in which $1$ must be Middle Square be covered twice and any of the Terminal squares of each row not even covered once. Now we will prove that in both the possibility Mentioned above. We can't have $6$ Vertical Strominos. Now just observe to have $6$ Vertical Strominos. There Must be at least one Column which must be covered by $2$ vertical Strominos. Now to do so we need to have middle square of that column not even covered once(as we also need $10$ horizontal strominos so this is only possible for the middle square of any of terminal column)$\implies$ That the $3^\text{rd}$ row is covered by $2^\text{nd}$ type. So inspite of Middle square of $3^\text{rd}$ row there is another square succsesive to it which is covered twice $\implies$ The column whose middle square is that particular column can't be covered by any Horizontal stromino. Then we need another One more Column with $2$ Horizontal strominos which is not possible so contradiction. Similarly we can achieve contradiction if those $10$ Strominos of $1$ kind will be Horizontal. Subcase 2-: Trying to cover the board by $9$ Strominos of $1$ kind and $7$ of another kind Prove-: Now take $9$ Strominos of $1$ kind (say Horizontal) then Exactly $4$ Rows will be covered By $2$ Horizontal Strominos and $1$ Row by $1$ Horizontal Stromino. And then as we need $7$ Vertical Strominos. So to make it happen We need at least $2$ Column Covered with $2$ Vertical Strominos. Then if for exactly $2$ Column it's Middle square is covered twice then Row $3$ Can't have any of Horizontal Stromino. Then we need another row which can have $3$ Horizontal Strominos which is not possible so contradiction. Similarly if we have $9$ Vertical strominos the we can also achieve contradiction by same process. Subcase 3-: When The board is covered by $8$ Strominos of $1$ Kind and $8$ of another kind. Prove -: Now take $8$ Horizontal and Vertical strominos. Then we will have Exactly $3$ Rows which will be covered by $2$ Horizontal Strominos and either of any $2$ Rows is not even covered by any Horizontal Stromino with $1$ row by $1$ Horizontal Stromino or both by $1$ Horizontal Stromino. And also to have $8$ Vertical Stromino We need similar Configuration of Covering Columns as in Row. So To have $3$ Columns Covered with $2$ Vertical Strominos. Then Clearly the Middle Most Row can't Be covered with any Stromino. SO all Rows(except Middle Row or $3^\text{rd}$ Row) Will be covered by $2$ Horizontal Strominos. And Then the The Middle most Column can't be covered by any Vertical column$\implies$ All other Column are Covered with $2$ Vertical Strominos. But then the Middle square of Middle Most column will be left uncovered Hence Contradiction. Hence, 16 Strominos can't cover the Every Square of Board satisfying given Conditions. $\blacksquare$

Color a table like this: How can I post attachment here? Since every stomino covers all 3 colors we see that only red colored cells can be covered with only one stomino. But by simmetry of table that can be only center cell. So we have only one cell which can be covered with only one stromino, but we need two. A contradiction.

Define a bad cell to be a square covered by exactly one stromino,a good cell its complement. Now Let $x_{i}$ be the number of strominos a cell is a part of.Now $\Sigma x_{i}=16*3$.We can conclude that there are exactly two bad cells. To locate the bad cells,we colour the board in two ways,the other being a $180$ anti-clockwise rotation of the first. Now,define the weight of the stromino as the sum of three numbers,it covers.We examine the first colouring first.Each stromino has weight $6$. Their total weight is $96$.The number of $1,2,3$ in the board are $9,8,8$.We do evaluations for when the bad cells are $(1,3),(2,2),(2,3),(1,2),(3,3)$. We get a contradiction in each of these cases by double counting the total weight.Thus both the bad cells are $1s$. Now,looking at the second colouring,we can similalry perform the same (as the number of each colors and weight of a stromino remain same),to get that only $1s$ can be bad.But combining both,we get that only the central square can be bad,which is a contradiciton.

@#18, Your solution is wrong because one of the 16 strominos will cover the remaining two squares resulting in $(23\cdot2 + 2)/3 = 16$ which is valid.

Suppose such a tiling of the board with $16$ strominos exist. Then we will double count the pairs ( cell $c$, stromino $S$ covering $c$). Let n be the number of cells that are covered by exactly one stromino. Then, $$n + 2(25-n) = 3\cdot 16$$which means $n=2$. Consider the following colouring of the board: \[\begin{bmatrix} 0&1&2&0&1\\ 1&2&0&1&2\\ 2&0&1&2&0\\ 0&1&2&0&1\\ 1&2&0&1&2 \end{bmatrix}\] There are $9$ squares that are numbered $1$, and $8$ squares that are numbered $0$ and $8$ squares numbered $2$ on them. Each stromino covers exactly one cell of each number. This means that the two squares that are covered exactly once must be the ones numbered $1$. Taking the mirror reflection of the above colouring we get: \[\begin{bmatrix} 1&0&2&1&0\\ 2&1&0&2&1\\ 0&2&1&0&2\\ 1&0&2&1&0\\ 2&1&0&2&1 \end{bmatrix}\]Applying the same reasoning from above, we get that the only squares that are covered once are the ones numbered $1$. Taking the intersection of both colorings we get: \[\begin{bmatrix} 1&0&2&1&0\\ 2&1&0&2&1\\ 0&2&\boxed{1}&0&2\\ 1&0&2&1&0\\ 2&1&0&2&1 \end{bmatrix}\]This means that $(3,3)$ is the only cell that is covered by exactly one stromino, contradicting the fact that there are two of them.$\blacksquare$

Huh... reading the problem wrong turned this into a five minute solve. The coloring idea here is identifying "key squares" in a manner very similar to this 高联 problem from 2007. Shade the squares in the middle row and the squares in the middle column, but excluding the center square. Every stromino covers at least one shaded square, and a stromino passing through the center square must cover at least two shaded squares. Now, the $8$ shaded squares must, in total, be covered at least $17$ times by any $16$-stromino covering, and we're done.