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 $A+P$ ?

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

NOTE: The original version of this problem was incorrect, as it was not stated that the side lengths must be positive integers, so $A+P$ could equal any of the answer choices. This problem was thrown out, and all contestants received full credit for it (whether they answered correctly, incorrectly, or left it blank). The problem above was modified to have an answer; as follows is the original 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 $A+P$ ?

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.

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 10 Problems and Solutions

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2015 AMC 10A Problems/Problem 20

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