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Is the square of 2013×2013 can be divided into rectangles 1×3 in such a way that the number of rectangles arranged vertically differ by 1 from the number rectangles arranged horizontally? Prove your answer

I totally have no idea how to solve that. I need accurate solutions or good hints.

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Try using 3x3 blocks to reduce the rectangle to a size you can reason about. –  Eric Tressler Oct 10 '13 at 19:29
    
Try determining the prime factors of 2013. –  stevemarvell Oct 10 '13 at 19:46
    
@stevemarvell 2013 = 3 · 11 · 61 –  user99183 Oct 10 '13 at 19:48
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1 Answer 1

up vote 3 down vote accepted

Since $2013$ is divisble by $3$, show by complete induction that the number of vertical bars that extend upward starting with bottom base at row $k$ must be divisible by $3$. Then you will get that the total number of vertical bars is divisible by $3$, and by symmetry, the number of horizontal bars that extend rightward starting with base column $k$ must be divisible by $3$, so the total number of horizontal bars must also be divisible by $3$, so they cannot differ by $1$.

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Ok, thank you, can I give me an example of induction ? Maybe formulas ? –  user99183 Oct 11 '13 at 14:17
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