CEMC Cayley

Since the rectangle has positive integer side lengths and an area of 24, its length and width must be a positive divisor pair of 24. Therefore, the length and width must be 24 and 1, or 12 and 2, or 8 and 3, or 6 and 4. Since the perimeter of a rectangle equals 2 times the sum of the length and width, the possible perimeters are $2(24+1)=50$, $2(12+2)=28$, $2(8+3)=22$, $2(6+4)=20$. These all appear as choices, which means that the perimeter of the rectangle cannot be 36.