2015 AMC 10A Problems/Problem 20

Problem

A rectangle has area $A$ $\text{cm}^2$ and perimeter $P$ $\text{cm}$ , where $A$ and $P$ are positive integers. 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 and width be $a$ and $b$ . Its area is $ab$ and the perimeter is $2(a + b)$ .

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

Looking at the answer choices, only $102$ cannot be written this way, because then either $a$ or $b$ would be $0$ .

So the answer is $\boxed{\textbf{(B) }102}$ .

Also, when adding 4 to 102, you get 106, which has fewer factors than 104, 108, 110, and 112.

Dispute

Unfortunately, the problem statement has an error: In order for the question to have a correct solution, the actual sides of the rectangle had to be integers. As stated, every answer choice would work, with one of the sides being $2$ , and the other, a half-integer. E.g., for $102$ , the sides of the rectangle would be $2$ and $49/2$ .

The AMC is allowing everyone 6 points for this question now.

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2015 AMC 10A Problems/Problem 20

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