Difference between revisions of "2002 AMC 8 Problems/Problem 11"
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==Solution 2== | ==Solution 2== | ||
| − | The edge of each square is one tile length longer than the edge of the previous square, which means that each square has <math>2*edge length - 1</math> more tiles than the previous square. Then the seventh square has <math>2*7-1=13</math> more tiles than the sixth square, which is <math>\boxed{\text{(C)}\ 13}</math>. | + | The edge of each square is one tile length longer than the edge of the previous square, which means that each square has <math>2*</math> edge length <math>- 1</math> more tiles than the previous square. Then the seventh square has <math>2*7-1=13</math> more tiles than the sixth square, which is <math>\boxed{\text{(C)}\ 13}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=10|num-a=12}} | {{AMC8 box|year=2002|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:59, 28 July 2013
Problem
A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?
![[asy] path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(3,0)*p); draw(shift(3,1)*p); draw(shift(4,0)*p); draw(shift(4,1)*p); draw(shift(7,0)*p); draw(shift(7,1)*p); draw(shift(7,2)*p); draw(shift(8,0)*p); draw(shift(8,1)*p); draw(shift(8,2)*p); draw(shift(9,0)*p); draw(shift(9,1)*p); draw(shift(9,2)*p);[/asy]](http://latex.artofproblemsolving.com/9/2/6/926b33e5dbe2d7e971f0ce8112e62c7a77b109a3.png)
Solution
Solution 1
The first square has a sidelength of 
, the second square 
, and so on. The seventh square has 
and is made of 
unit tiles. The sixth square has 
and is made of 
unit tiles. The seventh square has 
more tiles than the sixth square.
Solution 2
The edge of each square is one tile length longer than the edge of the previous square, which means that each square has 
edge length 
more tiles than the previous square. Then the seventh square has 
more tiles than the sixth square, which is 
.
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| 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. 
