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

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==Problem==
 
==Problem==
  
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A rectangle with positive integer side lengths has area <math>A</math> <math>\text{cm}^2</math> and perimeter <math>P</math> <math>\text{cm}</math>, where <math>A</math> and <math>P</math> are positive integers. Which of the following numbers cannot equal <math>A+P</math>?
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A rectangle has area <math>A</math> <math>\text{cm}^2</math> and perimeter <math>P</math> <math>\text{cm}</math>, where <math>A</math> and <math>P</math> are positive integers. Which of the following numbers cannot equal <math>A+P</math>?
  
 
<math> \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 </math>
 
<math> \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 </math>
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NOTE: As later announced by the MAA, this problem is incorrect, as it is not stated that the side lengths must be positive integers, so <math>A+P</math> can 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).
  
 
==Solution==
 
==Solution==

Revision as of 20:54, 5 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 $A+P$?

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

NOTE: As later announced by the MAA, this problem is incorrect, as it is not stated that the side lengths must be positive integers, so $A+P$ can 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).

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 original problem statement had an error: the problem had not mentioned that the original side lengths had to be positive integers. This caused every answer choice to 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 $49/2$. This problem has been modified correctly on the wiki for practicing uses.

The AMC gave everyone 6 points for this question.

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
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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