Difference between revisions of "2000 AMC 8 Problems/Problem 12"
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<math> \text{(A)}\ 344\qquad\text{(B)}\ 347\qquad\text{(C)}\ 350\qquad\text{(D)}\ 353\qquad\text{(E)}\ 356 </math> | <math> \text{(A)}\ 344\qquad\text{(B)}\ 347\qquad\text{(C)}\ 350\qquad\text{(D)}\ 353\qquad\text{(E)}\ 356 </math> | ||
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Since the bricks are <math>1</math> foot high, there will be <math>7</math> rows. To minimize the number of blocks used, rows <math>1, 3, 5,</math> and <math>7</math> will look like the bottom row of the picture, which takes <math>\frac{100}{2} = 50</math> bricks to construct. Rows <math>2, 4, </math> and <math>6</math> will look like the upper row pictured, which has <math>49</math> 2-foot bricks in the middle, and <math>2</math> 1-foot bricks on each end for a total of <math>51</math> bricks. | Since the bricks are <math>1</math> foot high, there will be <math>7</math> rows. To minimize the number of blocks used, rows <math>1, 3, 5,</math> and <math>7</math> will look like the bottom row of the picture, which takes <math>\frac{100}{2} = 50</math> bricks to construct. Rows <math>2, 4, </math> and <math>6</math> will look like the upper row pictured, which has <math>49</math> 2-foot bricks in the middle, and <math>2</math> 1-foot bricks on each end for a total of <math>51</math> bricks. | ||
| − | 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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| + | ==Video Solution by Soo== | ||
| + | https://youtu.be/EeHv96Ry_Ww - Soo, DRMS, NM | ||
==See Also== | ==See Also== | ||
Latest revision as of 23:55, 15 October 2023
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 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
