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>. | ||
| − | Also, when adding 4 to 102, you get 106, which has | + | Also, when adding 4 to 102, you get 106, which has fewer factors than 104, 108, 110, and 112. |
==No Solution== | ==No Solution== | ||
Revision as of 23:44, 4 February 2015
Contents
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 
.
Also, when adding 4 to 102, you get 106, which has fewer factors than 104, 108, 110, and 112.
No Solution
I am afraid the problem has an error: the actual sides of the rectangle had to be integers. As stated, every answer choice would work, with one of the sides being 
, and the other, a half-integer. E.g., for , the sides of the rectangle would be and 
.
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. 
