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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 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 with positive integer side lengths in <math>\text{cm}</math> has area <math>A</math> <math>\text{cm}^2</math> and perimeter <math>P</math> <math>\text{cm}</math>. 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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==Solution==
 
==Solution==
  
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Let the rectangle's length and width be <math>a</math> and <math>b</math>. Its area is <math>ab</math> and the perimeter is <math>2(a + b)</math>.
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Let the rectangle's length be <math>a</math> and its width be <math>b</math>. Its area is <math>ab</math> and the perimeter is <math>2a+2b</math>.
  
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Then <math>A + P = ab + 2a + 2b</math>. Factoring, this is <math>(a + 2)(b + 2) - 4</math>.
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Then <math>A + P = ab + 2a + 2b</math>. Factoring, we have <math>(a + 2)(b + 2) - 4</math>.
  
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Looking at the answer choices, only <math>102</math> cannot be written this way, because then either <math>a</math> or <math>b</math> would be <math>0</math>.
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The only one of the answer choices that cannot be expressed in this form is <math>102</math>, as <math>102 + 4</math> is twice a prime. There would then be no way to express <math>106</math> as <math>(a + 2)(b + 2)</math>, keeping <math>a</math> and <math>b</math> as positive integers.
  
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So the answer is <math>\boxed{\textbf{(B) }102}</math>.
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Our answer is then <math>\boxed{B}</math>.
  
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Also, when adding 4 to 102, you get 106, which has less factors than 104, 108, 110, and 112.
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Note: The original problem only stated that <math>A</math> and <math>P</math> were positive integers, not the side lengths themselves. This rendered the problem unsolvable, and so the AMC awarded everyone 6 points on this problem. This wiki has the corrected version of the problem so that the 2015 AMC 10A test can be used for practice.
  
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==No Solution==
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==Video Solution==
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https://youtu.be/RLo-e2On6Ac
  
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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>.
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~savannahsolver
  
 
==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}}
 +
 +
[[Category: Introductory Algebra Problems]]

Latest revision as of 19:16, 17 November 2020

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 $A+P$?

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

Solution

Let the rectangle's length be $a$ and its width be $b$. Its area is $ab$ and the perimeter is $2a+2b$.

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

The only one of the answer choices that cannot be expressed in this form is $102$, as $102 + 4$ is twice a prime. There would then be no way to express $106$ as $(a + 2)(b + 2)$, keeping $a$ and $b$ as positive integers.

Our answer is then $\boxed{B}$.

Note: The original problem only stated that $A$ and $P$ were positive integers, not the side lengths themselves. This rendered the problem unsolvable, and so the AMC awarded everyone 6 points on this problem. This wiki has the corrected version of the problem so that the 2015 AMC 10A test can be used for practice.

Video Solution

https://youtu.be/RLo-e2On6Ac

~savannahsolver

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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