Difference between revisions of "2015 AMC 10A Problems/Problem 20"
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The only one of the answer choices that cannot be expressed in this form is <math>102</math>, as <math>102 + 4</math> is twice a prime. There would then be no way to express <math>106</math> as <math>(a + 2)(b + 2)</math>, keeping <math>a</math> and <math>b</math> as positive integers. | The only one of the answer choices that cannot be expressed in this form is <math>102</math>, as <math>102 + 4</math> is twice a prime. There would then be no way to express <math>106</math> as <math>(a + 2)(b + 2)</math>, keeping <math>a</math> and <math>b</math> as positive integers. | ||
| − | Our answer is then <math>\boxed{B}</math> | + | Our answer is then <math>\boxed{B}</math>. |
Note: The original problem only stated that <math>A</math> and <math>P</math> were positive integers, not the side lengths themselves. This rendered the problem unsolvable, and so the AMC awarded everyone 6 points on this problem. This wiki has the corrected version of the problem so that the 2015 AMC 10A test can be used for practice. | Note: The original problem only stated that <math>A</math> and <math>P</math> were positive integers, not the side lengths themselves. This rendered the problem unsolvable, and so the AMC awarded everyone 6 points on this problem. This wiki has the corrected version of the problem so that the 2015 AMC 10A test can be used for practice. | ||
Revision as of 16:45, 1 January 2018
Problem
A rectangle with positive integer side lengths in 
has area 
and perimeter 
. Which of the following numbers cannot equal 
?
Solution
Let the rectangle's length be 
and its width be 
. Its area is 
and the perimeter is 
.
Then 
. Factoring, we have 
.
The only one of the answer choices that cannot be expressed in this form is 
, as 
is twice a prime. There would then be no way to express 
as 
, keeping and as positive integers.
Our answer is then 
.
Note: The original problem only stated that and were positive integers, not the side lengths themselves. This rendered the problem unsolvable, and so the AMC awarded everyone 6 points on this problem. This wiki has the corrected version of the problem so that the 2015 AMC 10A test can be used for practice.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
