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

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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 <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>. This problem has been modified correctly on the wiki for practicing uses.
 
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 <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>. This problem has been modified correctly on the wiki for practicing uses.
  
−
The AMC is giving everyone 6 points for this question.
+
The AMC gave everyone 6 points for this question.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=19|num-a=21}}
 
{{AMC10 box|year=2015|ab=A|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:33, 5 February 2015

Problem

A rectangle with positive integer side lengths 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 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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