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2015 AMC 10A Problem 20

2015 AMC 10A Problem 20AMC 10 Step by Step

Name: Date: October 28, 2026

Algebraic Manipulation · difficulty 3 of 5 · about 4 min · position P16-20

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 $A+P$ ?

A) $100$
B) $102$
C) $104$
D) $106$
E) $108$

Answer

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.

The trap

Hunting for rectangles by trial for each choice and guessing when none turns up, with no reason why 102 is really impossible.

Concise solution

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)

Techniques

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