Difference between revisions of "2015 AMC 10A Problems/Problem 20"
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So the answer is <math>\boxed{\textbf{(B) }102}</math>. | So the answer is <math>\boxed{\textbf{(B) }102}</math>. | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=19|num-a=21}} | {{AMC10 box|year=2015|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:49, 9 February 2015
Problem
A rectangle with positive integer side lengths in 
has area 
and perimeter 
. Which of the following numbers cannot equal 
?
NOTE: The original version of this problem was incorrect, as it was not stated that the side lengths must be positive integers, so could equal any of the answer choices. This problem was thrown out, and all contestants received full credit for it (whether they answered correctly, incorrectly, or left it blank). The problem above was modified to have an answer; as follows is the original problem:
A rectangle has area and perimeter , where and are positive integers. Which of the following numbers cannot equal ?
Solution
Let the rectangle's length and width be 
and 
. Its area is 
and the perimeter is 
.
Then 
. Factoring, this is 
.
Looking at the answer choices, only 
cannot be written this way, because then either or would be 
.
So the answer is 
.
See Also
| 2015 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
