Problem 271

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, which is (E).