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

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Also, when adding 4 to 102, you get 106, which has less factors than 104, 108, 110, and 112.
 
Also, when adding 4 to 102, you get 106, which has less factors than 104, 108, 110, and 112.
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==No Solution==
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I am afraid the problem has an error: the actual sides of the rectangle had to be integers.  As stated, every answer choice would work, with one of the sides being <math>2</math>, and the other, a half-integer.  E.g., for <math>102</math>, the sides of the rectangle would be <math>2</math> and <math>49/2</math>.

Revision as of 20:07, 4 February 2015

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 less factors than 104, 108, 110, and 112.

No Solution

I am afraid the problem has an error: 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 $49/2$.

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