2015 AMC 10A Problem 20

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

An expression of the form $xy + ax + by$ becomes a product $(x + b)(y + a)$ after adding the constant $ab$; that is Simon's Favorite Factoring Trick. It converts "can this value be achieved?" into "does this shifted number have a factorization with the right constraints?", which is a quick divisibility check. Whenever a problem mixes a product of two unknowns with linear terms in the same unknowns, try completing the rectangle.

Reason about the shifted numbers directly: $N + 4$ must have a divisor $d$ with $3 \le d \le \sqrt{N + 4}$. Among $104, 106, 108, 110, 112$, only $106$ is twice a prime, so it alone has no such divisor.

• Hunting for a rectangle for each choice by trial and stopping when one is not found, with no factorization to prove $102$ is impossible.
• Allowing a factor of $2$, i.e. a side of length $0$, which would make $106 = 2 \cdot 53$ appear to work.
• Factoring the choices themselves instead of the choices plus $4$; for example $102 = 6 \cdot 17$ has nothing to do with the problem.