2015 AMC 10A Problems/Problem 20

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.

2015 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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2015 AMC 10A Problems/Problem 20

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