2022 CEMC Cayley Problems/Problem 8

Problem

A rectangle has positive integer side lengths and an area of $24$ . The perimeter of the rectangle cannot be

$\text{ (A) }\ 20 \qquad\text{ (B) }\ 22 \qquad\text{ (C) }\ 28 \qquad\text{ (D) }\ 50 \qquad\text{ (E) }\ 36$

Solution 1

The area of the rectangle is just its side lengths multiplied together, so we need to look at factor pairs for $24$ because we know that both of the side lengths are integers.

The possible factor pairs (excluding flipping the numbers) for $24$ are $(1, 24)$ , $(2, 12)$ , $(3, 8)$ , and $(4, 6)$ .

If one side length is $1$ and the other side length is $24$ , then the perimeter is $2(1 + 24) = 2 \times 25 = 50$ . This eliminates answer choice D, as it is possible to achieve.

Using the other lengths and the same process for each of them, we get perimeters of $2(2 + 12) = 2 \times 14 = 28$ , $2(3 + 8) = 2 \times 11 = 22$ , and $2(4 + 6) = 2 \times 10 = 20$ . This eliminates answer choices A, B, and C.

Because all of the other answer choices have been eliminated, we know that the answer is $\boxed {\textbf {(E) } 36}$ .

~anabel.disher

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2022 CEMC Cayley Problems/Problem 8

Problem

Solution 1