1998 AHSME Problems/Problem 1

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Problem 1

Each of the sides of five congruent rectangles is labeled with an integer. In rectangle A, $w = 4, x = 1, y = 6, z = 9$. In rectangle B, $w = 1, x = 0, y = 3, z = 6$. In rectangle C, $w = 3, x = 8, y = 5, z = 2$. In rectangle D, $w = 7, x = 5, y = 4, z = 8$. In rectangle E, $w = 9, x = 2, y = 7, z = 0$. These five rectangles are placed, without rotating or reflecting, in position as below. Which of the rectangle is the top leftmost one?

[asy] draw((0,5)--(0,7)--(3,7)--(3,5)--cycle); draw((0,4)--(9,4)--(9,2)--(6,2)--(6,0)--(0,0)--cycle); draw((3,0)--(3,4));draw((6,2)--(6,4));draw((0,2)--(6,2)); label("$w$",(0,6),(-1,0));label("$x$",(1.5,7),(0,1));label("$y$",(3,6),(1,0));label("$z$",(1.5,5),(0,-1)); [/asy]

$\mathrm{(A)\ } A \qquad \mathrm{(B) \ }B \qquad \mathrm{(C) \  } C \qquad \mathrm{(D) \  } D \qquad \mathrm{(E) \  }E$

Solution

Looking at the list of $w$ and $y$ values that must match up left-to-right, we have $(w,y) = A(4,6), B(1,3), C(3,5), D(7,4), E(9,7)$. Looking for digits that only appear once, we see that $9$, $6$, $1$, and $5$ cannot match up to other digits, and thus must appear on the ends. $1$ and $9$ only appear on the left, and thus their respective blocks $B$ and $E$ must appear on the left. Similarly, $6$ and $5$ only appear on the right, and thus their blocks $A$ and $C$ must appear on the right-most block of their row. Therefore, $D$, the only block without a unique digit, must be the top-center block.

Now that we have placed one block, $D$, we can only place block $E$ to the left of $D$, and block $A$ to the right of $D$. Thus, $\boxed{E}$ is the right answer.

Completing the puzzle, the top boxes read $EDA$, while the bottom two boxes read $BC$.


See Also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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