Difference between revisions of "2015 AMC 10A Problems/Problem 20"
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| − | Let the | + | Let the square's length and width be <math>a</math> . Its area is <math>a^2</math> and the perimeter is <math>2(a^2)</math>. |
| − | Then <math>A + P = | + | Then <math>A + P = a^2+ 4a</math>. Factoring, this is <math>(a + 2)(a + 2) - 4</math>. |
| − | Looking at the answer choices, only <math>102</math> cannot be written this way, because then | + | Looking at the answer choices, only <math>102</math> cannot be written this way, because then <math>a</math> would be <math>0</math>. |
So the answer is <math>\boxed{\textbf{(B) }102}</math>. | So the answer is <math>\boxed{\textbf{(B) }102}</math>. | ||
Revision as of 17:22, 31 January 2016
Problem
A square with positive integer side lengths in 
has area 
and perimeter 
. Which of the following numbers cannot equal 
?
Solution
Let the square's length and width be 
. Its area is 
and the perimeter is 
.
Then 
. Factoring, this is 
.
Looking at the answer choices, only 
cannot be written this way, because then 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. 
