Difference between revisions of "2008 AMC 12B Problems/Problem 16"
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<math>8=ab-4a-4b+16=(a-4)(b-4)</math> | <math>8=ab-4a-4b+16=(a-4)(b-4)</math> | ||
| − | Since <math>8=1 | + | Since <math>8=1\times8</math> and <math>8=2\times4</math> are the only positive factorings of <math>8</math>. |
<math>(a,b)=(5,12)</math> or <math>(a,b)=(6,8)</math> yielding <math>2\Rightarrow \textbf{(B)}</math> solutions. Notice that because <math>b>a</math>, the reversed pairs are invalid. | <math>(a,b)=(5,12)</math> or <math>(a,b)=(6,8)</math> yielding <math>2\Rightarrow \textbf{(B)}</math> solutions. Notice that because <math>b>a</math>, the reversed pairs are invalid. | ||
Revision as of 11:57, 2 February 2015
Problem
A rectangular floor measures 
by 
feet, where and are positive integers with 
. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width 
foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair 
?
Solution
By Simon's Favorite Factoring Trick:
Since 
and 
are the only positive factorings of 
.
or 
yielding 
solutions. Notice that because 
, the reversed pairs are invalid.
See Also
| 2008 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
