Difference between revisions of "2008 AMC 12B Problems/Problem 16"
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Since <math>8=1\times8</math> and <math>8=2\times4</math> are the only positive factorings of <math>8</math>. | 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> | + | <math>(a,b)=(5,12)</math> or <math>(a,b)=(6,8)</math> yielding <math>\Rightarrow\textbf{(B)}</math> <math>2</math> solutions. Notice that because <math>b>a</math>, the reversed pairs are invalid. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2008|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:03, 12 October 2021
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.