Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AMC 10A Problems/Problem 20"

(→‎Solution)
 
(6 intermediate revisions by 4 users not shown)
Line 11: Line 11:
 
Then <math>A + P = ab + 2a + 2b</math>. Factoring, we have <math>(a + 2)(b + 2) - 4</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 connect 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.
 +
 +
==Video Solution==
 +
https://youtu.be/RLo-e2On6Ac
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 19:16, 17 November 2020

Problem

A rectangle with positive integer side lengths in $\text{cm}$ has area $A$ $\text{cm}^2$ and perimeter $P$ $\text{cm}$. Which of the following numbers cannot equal $A+P$?

$\textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108$

Solution

Let the rectangle's length be $a$ and its width be $b$. Its area is $ab$ and the perimeter is $2a+2b$.

Then $A + P = ab + 2a + 2b$. Factoring, we have $(a + 2)(b + 2) - 4$.

The only one of the answer choices that cannot be expressed in this form is $102$, as $102 + 4$ is twice a prime. There would then be no way to express $106$ as $(a + 2)(b + 2)$, keeping $a$ and $b$ as positive integers.

Our answer is then $\boxed{B}$.

Note: The original problem only stated that $A$ and $P$ 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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_20&oldid=137214"