View source for 2015 AMC 10A Problems/Problem 20

==Problem==

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>?

<math> \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 </math>

==Solution==

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>.

Then <math>A + P = ab + 2a + 2b</math>. Factoring, this is <math>(a + 2)(b + 2) - 4</math>.

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>.

So the answer is <math>\boxed{\textbf{(B) }102}</math>.

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 <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.

==See Also==
{{AMC10 box|year=2015|ab=A|num-b=19|num-a=21}}
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