2015 AMC 10A Problems/Problem 20

Problem

A square 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 $2(a^2)$ .

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 $\boxed{\textbf{(B) }102}$ .

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.

Page

Toolbox

Search

2015 AMC 10A Problems/Problem 20

Problem

Solution

See Also