2015 AMC 10A Problems/Problem 20

Revision as of 14:28, 1 February 2016 by DeathLlama9 (talk | contribs) (it's not a square so)

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 square's length and width be $a$ . Its area is $a^2$ and the perimeter is $4a$.

Then $A + P = a^2+ 4a$. Factoring, this is $(a + 2)(a + 2) - 4$.

Looking at the answer choices, only $102$ cannot be written this way, because then $a$ would be $0$.

So the answer is $\boxed{\textbf{(B) }102}$.

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
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