2007 AMC 8 Problems/Problem 11

Problem

Tiles $I, II, III$ and $IV$ are translated so one tile coincides with each of the rectangles $A, B, C$ and $D$. In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle $C$?

[asy] size(400); defaultpen(linewidth(0.8)); path p=origin--(8,0)--(8,6)--(0,6)--cycle; draw(p^^shift(8.5,0)*p^^shift(8.5,10)*p^^shift(0,10)*p); draw(shift(20,2)*p^^shift(28,2)*p^^shift(20,8)*p^^shift(28,8)*p); label("8", (4,6+10), S); label("6", (4+8.5,6+10), S); label("7", (4,6), S); label("2", (4+8.5,6), S); label("I", (4,6+10), N); label("II", (4+8.5,6+10), N); label("III", (4,6), N); label("IV", (4+8.5,6), N); label("3", (0,3+10), E); label("4", (0+8.5,3+10), E); label("1", (0,3), E); label("9", (0+8.5,3), E); label("7", (4,10), N); label("2", (4+8.5,10), N); label("0", (4,0), N); label("6", (4+8.5,0), N); label("9", (8,3+10), W); label("3", (8+8.5,3+10), W); label("5", (8,3), W); label("1", (8+8.5,3), W); label("A", (24,10), N); label("B", (32,10), N); label("C", (24,4), N); label("D", (32,4), N);[/asy]

$\mathrm{(A)}\ I \qquad \mathrm{(B)}\ II \qquad \mathrm{(C)}\ III \qquad \mathrm{(D)}\ IV \qquad \mathrm{(E)}$ cannot be determined

Solution

We first notice that tile III has a $0$ on the bottom and a $5$ on the right side. Since no other tile has a $0$ or a $5$, Tile III must be in rectangle $D$. Tile III also has a $1$ on the left, so Tile IV must be in Rectangle $C$.

The answer is $\boxed{\textbf{(D)}}$

AMC8 2007 11S.png

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS