Can a $10times 10$ square be entirely covered by 25 $T$-shape bricks?

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Let $ABCD$ be a square in which length of a side is $10$ meters. Suppose that we have $T$-shape brick which consists of $4$ smaller squares in which a side of each smaller square has length of $1$ meter. Can $ABCD$ be entirely covered by 25 $T$-shape bricks?



I have tried but can not figure out where to start. Please give me some hints, not the full solution!



enter image description here




combinatorics recreational-mathematics polyomino




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  • 4




    Use colouring proofs.
    – Anik Bhowmick
    Aug 6 at 7:12






  • 2




    If the grid is coloured like a checkerboard, $25$ T-shapes can never cover the same number of black squares as red.
    – Robert Israel
    Aug 6 at 7:20






  • 1




    Closely related (but not a dupe!)
    – Jyrki Lahtonen
    Aug 6 at 7:36














up vote
4
down vote

favorite












Let $ABCD$ be a square in which length of a side is $10$ meters. Suppose that we have $T$-shape brick which consists of $4$ smaller squares in which a side of each smaller square has length of $1$ meter. Can $ABCD$ be entirely covered by 25 $T$-shape bricks?



I have tried but can not figure out where to start. Please give me some hints, not the full solution!



enter image description here




combinatorics recreational-mathematics polyomino




share|cite|improve this question

















  • 4




    Use colouring proofs.
    – Anik Bhowmick
    Aug 6 at 7:12






  • 2




    If the grid is coloured like a checkerboard, $25$ T-shapes can never cover the same number of black squares as red.
    – Robert Israel
    Aug 6 at 7:20






  • 1




    Closely related (but not a dupe!)
    – Jyrki Lahtonen
    Aug 6 at 7:36












up vote
4
down vote

favorite









up vote
4
down vote

favorite











Let $ABCD$ be a square in which length of a side is $10$ meters. Suppose that we have $T$-shape brick which consists of $4$ smaller squares in which a side of each smaller square has length of $1$ meter. Can $ABCD$ be entirely covered by 25 $T$-shape bricks?



I have tried but can not figure out where to start. Please give me some hints, not the full solution!



enter image description here




combinatorics recreational-mathematics polyomino




share|cite|improve this question













Let $ABCD$ be a square in which length of a side is $10$ meters. Suppose that we have $T$-shape brick which consists of $4$ smaller squares in which a side of each smaller square has length of $1$ meter. Can $ABCD$ be entirely covered by 25 $T$-shape bricks?



I have tried but can not figure out where to start. Please give me some hints, not the full solution!



enter image description here





combinatorics recreational-mathematics polyomino





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 6 at 12:58









Herman Tulleken

786417




786417









asked Aug 6 at 7:05









Le Anh Dung

723318




723318







  • 4




    Use colouring proofs.
    – Anik Bhowmick
    Aug 6 at 7:12






  • 2




    If the grid is coloured like a checkerboard, $25$ T-shapes can never cover the same number of black squares as red.
    – Robert Israel
    Aug 6 at 7:20






  • 1




    Closely related (but not a dupe!)
    – Jyrki Lahtonen
    Aug 6 at 7:36












  • 4




    Use colouring proofs.
    – Anik Bhowmick
    Aug 6 at 7:12






  • 2




    If the grid is coloured like a checkerboard, $25$ T-shapes can never cover the same number of black squares as red.
    – Robert Israel
    Aug 6 at 7:20






  • 1




    Closely related (but not a dupe!)
    – Jyrki Lahtonen
    Aug 6 at 7:36







4




4




Use colouring proofs.
– Anik Bhowmick
Aug 6 at 7:12




Use colouring proofs.
– Anik Bhowmick
Aug 6 at 7:12




2




2




If the grid is coloured like a checkerboard, $25$ T-shapes can never cover the same number of black squares as red.
– Robert Israel
Aug 6 at 7:20




If the grid is coloured like a checkerboard, $25$ T-shapes can never cover the same number of black squares as red.
– Robert Israel
Aug 6 at 7:20




1




1




Closely related (but not a dupe!)
– Jyrki Lahtonen
Aug 6 at 7:36




Closely related (but not a dupe!)
– Jyrki Lahtonen
Aug 6 at 7:36










1 Answer
1






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up vote
7
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Think to the given square as a $10times 10$ chessboard with alternate black and white squares and assume that such covering with $T$ pieces exists.
Then each of the $25$ $T$ pieces will have 1) $3$ black squares and $1$ white square or 2) $3$ white squares and $1$ black square. Let $b$ the number of $T$ pieces of the first category and $w$ the number of $T$ pieces of the second category. Then the integers $b$ and $w$ should satisfy the following equations:
$$3b+w=frac10cdot 102,quad b+3w=frac10cdot 102.$$
What may we conclude?



Bonus question: What happens when the given square is $ntimes n$?






share|cite|improve this answer























  • Do you know the answer for the $m$-by-$n$ rectangular case? It suffices that both $m$ and $n$ are divisible by $4$, but I cannot prove that this is necessary. I know that one of $m$ and $n$ is divisible by $4$ (or better, just as you showed, $8mid mn$).
    – Batominovski
    Aug 6 at 8:14







  • 1




    @Batominovski Yes, both $m$ and $n$ are divisible by 4 is also necessary. See jstor.org/stable/pdf/2313337.pdf?seq=1#page_scan_tab_contents
    – Robert Z
    Aug 6 at 8:30










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up vote
7
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Think to the given square as a $10times 10$ chessboard with alternate black and white squares and assume that such covering with $T$ pieces exists.
Then each of the $25$ $T$ pieces will have 1) $3$ black squares and $1$ white square or 2) $3$ white squares and $1$ black square. Let $b$ the number of $T$ pieces of the first category and $w$ the number of $T$ pieces of the second category. Then the integers $b$ and $w$ should satisfy the following equations:
$$3b+w=frac10cdot 102,quad b+3w=frac10cdot 102.$$
What may we conclude?



Bonus question: What happens when the given square is $ntimes n$?






share|cite|improve this answer























  • Do you know the answer for the $m$-by-$n$ rectangular case? It suffices that both $m$ and $n$ are divisible by $4$, but I cannot prove that this is necessary. I know that one of $m$ and $n$ is divisible by $4$ (or better, just as you showed, $8mid mn$).
    – Batominovski
    Aug 6 at 8:14







  • 1




    @Batominovski Yes, both $m$ and $n$ are divisible by 4 is also necessary. See jstor.org/stable/pdf/2313337.pdf?seq=1#page_scan_tab_contents
    – Robert Z
    Aug 6 at 8:30














up vote
7
down vote













Think to the given square as a $10times 10$ chessboard with alternate black and white squares and assume that such covering with $T$ pieces exists.
Then each of the $25$ $T$ pieces will have 1) $3$ black squares and $1$ white square or 2) $3$ white squares and $1$ black square. Let $b$ the number of $T$ pieces of the first category and $w$ the number of $T$ pieces of the second category. Then the integers $b$ and $w$ should satisfy the following equations:
$$3b+w=frac10cdot 102,quad b+3w=frac10cdot 102.$$
What may we conclude?



Bonus question: What happens when the given square is $ntimes n$?






share|cite|improve this answer























  • Do you know the answer for the $m$-by-$n$ rectangular case? It suffices that both $m$ and $n$ are divisible by $4$, but I cannot prove that this is necessary. I know that one of $m$ and $n$ is divisible by $4$ (or better, just as you showed, $8mid mn$).
    – Batominovski
    Aug 6 at 8:14







  • 1




    @Batominovski Yes, both $m$ and $n$ are divisible by 4 is also necessary. See jstor.org/stable/pdf/2313337.pdf?seq=1#page_scan_tab_contents
    – Robert Z
    Aug 6 at 8:30












up vote
7
down vote










up vote
7
down vote









Think to the given square as a $10times 10$ chessboard with alternate black and white squares and assume that such covering with $T$ pieces exists.
Then each of the $25$ $T$ pieces will have 1) $3$ black squares and $1$ white square or 2) $3$ white squares and $1$ black square. Let $b$ the number of $T$ pieces of the first category and $w$ the number of $T$ pieces of the second category. Then the integers $b$ and $w$ should satisfy the following equations:
$$3b+w=frac10cdot 102,quad b+3w=frac10cdot 102.$$
What may we conclude?



Bonus question: What happens when the given square is $ntimes n$?






share|cite|improve this answer















Think to the given square as a $10times 10$ chessboard with alternate black and white squares and assume that such covering with $T$ pieces exists.
Then each of the $25$ $T$ pieces will have 1) $3$ black squares and $1$ white square or 2) $3$ white squares and $1$ black square. Let $b$ the number of $T$ pieces of the first category and $w$ the number of $T$ pieces of the second category. Then the integers $b$ and $w$ should satisfy the following equations:
$$3b+w=frac10cdot 102,quad b+3w=frac10cdot 102.$$
What may we conclude?



Bonus question: What happens when the given square is $ntimes n$?







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Aug 6 at 7:43


























answered Aug 6 at 7:26









Robert Z

84.2k955123




84.2k955123











  • Do you know the answer for the $m$-by-$n$ rectangular case? It suffices that both $m$ and $n$ are divisible by $4$, but I cannot prove that this is necessary. I know that one of $m$ and $n$ is divisible by $4$ (or better, just as you showed, $8mid mn$).
    – Batominovski
    Aug 6 at 8:14







  • 1




    @Batominovski Yes, both $m$ and $n$ are divisible by 4 is also necessary. See jstor.org/stable/pdf/2313337.pdf?seq=1#page_scan_tab_contents
    – Robert Z
    Aug 6 at 8:30
















  • Do you know the answer for the $m$-by-$n$ rectangular case? It suffices that both $m$ and $n$ are divisible by $4$, but I cannot prove that this is necessary. I know that one of $m$ and $n$ is divisible by $4$ (or better, just as you showed, $8mid mn$).
    – Batominovski
    Aug 6 at 8:14







  • 1




    @Batominovski Yes, both $m$ and $n$ are divisible by 4 is also necessary. See jstor.org/stable/pdf/2313337.pdf?seq=1#page_scan_tab_contents
    – Robert Z
    Aug 6 at 8:30















Do you know the answer for the $m$-by-$n$ rectangular case? It suffices that both $m$ and $n$ are divisible by $4$, but I cannot prove that this is necessary. I know that one of $m$ and $n$ is divisible by $4$ (or better, just as you showed, $8mid mn$).
– Batominovski
Aug 6 at 8:14





Do you know the answer for the $m$-by-$n$ rectangular case? It suffices that both $m$ and $n$ are divisible by $4$, but I cannot prove that this is necessary. I know that one of $m$ and $n$ is divisible by $4$ (or better, just as you showed, $8mid mn$).
– Batominovski
Aug 6 at 8:14





1




1




@Batominovski Yes, both $m$ and $n$ are divisible by 4 is also necessary. See jstor.org/stable/pdf/2313337.pdf?seq=1#page_scan_tab_contents
– Robert Z
Aug 6 at 8:30




@Batominovski Yes, both $m$ and $n$ are divisible by 4 is also necessary. See jstor.org/stable/pdf/2313337.pdf?seq=1#page_scan_tab_contents
– Robert Z
Aug 6 at 8:30












 

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