Difference between revisions of "2015 AMC 10A Problems/Problem 20"
(13 intermediate revisions by 8 users not shown) | |||
Line 5: | Line 5: | ||
<math> \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 </math> | <math> \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 </math> | ||
− | + | ==Solution== | |
− | + | Let the rectangle's length be <math>a</math> and its width be <math>b</math>. Its area is <math>ab</math> and the perimeter is <math>2a+2b</math>. | |
− | = | + | Then <math>A + P = ab + 2a + 2b</math>. Factoring, we have <math>(a + 2)(b + 2) - 4</math>. |
+ | |||
+ | 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>. | |
− | + | 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. | |
− | + | ==Video Solution== | |
+ | https://youtu.be/RLo-e2On6Ac | ||
− | + | ~savannahsolver | |
==See Also== | ==See Also== |
Latest revision as of 20:16, 17 November 2020
Contents
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.
Video Solution
~savannahsolver
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.