2003 AMC 8 Problems/Problem 13: Difference between revisions
No edit summary |
No edit summary |
||
| Line 30: | Line 30: | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2003|num-b=12|num-a=14}} | {{AMC8 box|year=2003|num-b=12|num-a=14}} | ||
{{MAA Notice}} | |||
Revision as of 23:47, 4 July 2013
Problem
Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?
![[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight); [/asy]](http://latex.artofproblemsolving.com/a/d/8/ad8d99504ec6b48d8afd6b366c02c827e11c198e.png)
Solution
This is the number cubes that are adjacent to another cube on two sides. The bottom corner cubes are connected on three sides, and the top corner cubes are connected on one. The number we are looking for is the number of middle cubes, which is 
.
See Also
| 2003 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
