2015 AMC 10A Problem 20

P16-20~4 minPrint

Algebraic Manipulation Diophantine Equations Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Use the answer choices (mod checks, size, form) to eliminate or select

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 $A+P$ ?

Show solution without answering

Key insight

Area plus perimeter plus 4 always factors as (width + 2) times (length + 2), both at least 3. But 106 is only 2 times 53.

Solution

Find a shape for $A + P$ that is easy to test, then check the choices.

1. Let the sides be whole numbers $x$ and $y$ . The area is $xy$ and the perimeter is $2x + 2y$ , so
$A + P = xy + 2x + 2y.$
2. This almost factors. Look at what $(x + 2)(y + 2)$ multiplies out to: $xy + 2x + 2y + 4$ . That is our expression plus $4$ . So
$A + P + 4 = (x + 2)(y + 2).$
3. Since $x$ and $y$ are at least $1$ , both $x + 2$ and $y + 2$ are at least $3$ . So a number can be $A + P$ exactly when that number plus $4$ can be written as a product of two whole numbers that are each $3$ or more.
4. Test each choice:

choice	plus $4$	split	rectangle
$100$	$104$	$4 \times 26$	$2 \times 24$
$102$	$106$	only $1 \times 106$ or $2 \times 53$	none
$104$	$108$	$9 \times 12$	$7 \times 10$
$106$	$110$	$10 \times 11$	$8 \times 9$
$108$	$112$	$8 \times 14$	$6 \times 12$

The number $53$ is prime, so $106$ has no other splits. Both possible splits use a factor smaller than $3$ . So no rectangle gives $102$ .
Check one that works: a $7 \times 10$ rectangle has area $70$ and perimeter $34$ , and $70 + 34 = 104$ .

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

Why this works

An expression like $xy + 2x + 2y$ becomes a product once you add the right constant, here $2 \times 2 = 4$ . People call this Simon's Favorite Factoring Trick. It changes "can we find a rectangle?" into "does this number have the right factors?", which is quick to check. Try it whenever a product of two unknowns appears together with each unknown on its own.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let the sides be positive integers $x$ and $y$ . Then
$A + P = xy + 2x + 2y.$
Adding $4$ completes a product:
$A + P + 4 = xy + 2x + 2y + 4 = (x + 2)(y + 2).$
Since $x, y \ge 1$ , both factors are at least $3$ . So a number $N$ can equal $A + P$ exactly when $N + 4$ splits as a product of two integers that are each at least $3$ .

Test the choices:

$100 + 4 = 104 = 4 \cdot 26$ , so $x = 2$ , $y = 24$ works.
$102 + 4 = 106 = 2 \cdot 53$ . Because $53$ is prime, the only factorizations are $1 \cdot 106$ and $2 \cdot 53$ , and each has a factor smaller than $3$ . No rectangle exists.
$104 + 4 = 108 = 9 \cdot 12$ , giving a $7 \times 10$ rectangle.
$106 + 4 = 110 = 10 \cdot 11$ , giving $8 \times 9$ .
$108 + 4 = 112 = 8 \cdot 14$ , giving $6 \times 12$ .

Only $102$ is impossible. The answer is $\boxed{\textbf{(B)}\ 102}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Algebra, Chapter 22: Special Manipulations
Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.4)