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

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Our answer is then <math>\boxed{B}</math>
 
Our answer is then <math>\boxed{B}</math>
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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.
  
 
==See Also==
 
==See Also==

Revision as of 00:12, 5 February 2016

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 connect 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.

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
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

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