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Blue and red rectangles are drawn on a blackboard. Exactly 7 of the rectangles are squares. There are 3 red rectangles more than blue squares. There are 2 red squares more than blue rectangles. How many blue rectangles are there?

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    $\begingroup$ What did you tried to solve the problem? $\endgroup$ – choco_addicted Feb 4 '16 at 8:20
  • $\begingroup$ I tried to solve the question by writing the statements in the forms of equations like Rs+Bs=7 and solve them simultaneously.However,after a few non-succesful attempts and realizing that there are 4 unknowns but only 3 equations,i gave up. $\endgroup$ – Max Feb 4 '16 at 10:02
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Let $bs$,$bn$,$rs$, and $rn$ denote the number of blue squares, blue rectangles that are not squares, red squares and red rectangles that are not squares respectively, where each must be a non negative integer.

Then, the given conditions give the following equations: $$rs+bs=7$$ $$bs+3=rs+rn$$ $$bs+bn+2=rs$$

Using the first and the third equation to eliminate $rs$, we get

$$bs+bn+2=7-bs$$ or $$2bs+bn=5$$ Now, $$rs=7-bs$$ $$rs+rn=bs+3\ge rs$$ Thus $$bs+3\ge 7-bs$$ $$bs\ge 2$$ For $bs$ and $bn$ to be non negative integers, in $2bs+bn=5$, we have $bs\le2$

Thus $$bs=2$$ $$bn=5-2bs=1$$ Also $rs=5$ and $rn=0$

The number of blue rectangles is $bs+bn=3$

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Let $b,b',r,r'$ be

  • $b$: the number of blue rectangles which are not square
  • $b'$: the number of blue squares
  • $r$: the number of red rectangles which are not square
  • $r'$: the number of red squares.

From "Exactly 7 of the rectangles are squares.", we know $b'+r'=7$, from "There are 3 red rectangles more than blue squares.", we know $r+r'=3+b'$, and from "There are 2 red squares more than blue rectangles.", we know $b'=2+b+b'$. Therefore, we can set the system of linear equations $$\left\{\begin{array}{ccccccccc} 0&+&b'&+&0&+&r'&=&7\\ 0&-&b'&+&r&+&r'&=&3\\ b&+&b'&+&0&-&r'&=&-2 \end{array}\right.$$ Using Gaussian elimination, we get

\begin{cases} b&=&-9&+&2r'\\ b'&=&7&-&r'\\ r&=&10&-&2r' \end{cases} Since $b,b',r,r'$ must be nonnegative integers, $r'$ is $5$. Therefore, $b+b'=1+2=3$.

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  • $\begingroup$ Thx,but what is Gaussian elimination?How is it done? $\endgroup$ – Max Feb 4 '16 at 10:18
  • $\begingroup$ Gaussian elimination is a method to solve linear system and I learned it in a linear algebra course. For detailed description, see here. $\endgroup$ – choco_addicted Feb 4 '16 at 10:27
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Let there be $b$ blue and $r$ red rectangles, among them $b'$ and $r'$ squares. According to the text we have in turn $$r'=2+b,\quad b'=7-r'=5-b,\quad r=b'+3=8-b\ .\tag{1}$$ The conditions $r'\leq r$ and $b'\leq b$ enforce $2b\leq6$ and $5\leq2b$, so that necessarily $b=3$. From $(1)$ we then obtain $r'=r=5$, $b'=2$.

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