Difference between revisions of "2000 AMC 8 Problems/Problem 12"
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Four rows of <math>50</math> bricks and three rows of <math>51</math> bricks totals <math>4\cdot 50 + 3\cdot 51 = 200 + 153 = 353</math> bricks, giving the answer <math>\boxed{D}.</math> | Four rows of <math>50</math> bricks and three rows of <math>51</math> bricks totals <math>4\cdot 50 + 3\cdot 51 = 200 + 153 = 353</math> bricks, giving the answer <math>\boxed{D}.</math> | ||
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| + | If you wanted to be extra sure, you can realize that <math>700 / 2</math> = 350 blocks. However, this is impossible, as you need them to be staggered. | ||
==Video Solution by Soo== | ==Video Solution by Soo== | ||
Latest revision as of 17:56, 6 September 2025
Problem
A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long (no blocks may be cut). The vertical joins in the blocks must be staggered as shown, and the wall must be even on the ends. What is the smallest number of blocks needed to build this wall?
![[asy] draw((0,0)--(6,0)--(6,1)--(5,1)--(5,2)--(0,2)--cycle); draw((0,1)--(5,1)); draw((1,1)--(1,2)); draw((3,1)--(3,2)); draw((2,0)--(2,1)); draw((4,0)--(4,1)); [/asy]](http://latex.artofproblemsolving.com/0/4/e/04ee50ee4b283b079f7a9cd1f2084ff60439b123.png)
Solution
Since the bricks are 
foot high, there will be 
rows. To minimize the number of blocks used, rows 
and will look like the bottom row of the picture, which takes 
bricks to construct. Rows 
and 
will look like the upper row pictured, which has 
2-foot bricks in the middle, and 
1-foot bricks on each end for a total of 
bricks.
Four rows of 
bricks and three rows of bricks totals 
bricks, giving the answer 
If you wanted to be extra sure, you can realize that 
= 350 blocks. However, this is impossible, as you need them to be staggered.
Video Solution by Soo
https://youtu.be/EeHv96Ry_Ww - Soo, DRMS, NM
See Also
| 2000 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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, as part of the American Mathematics Competitions. 
