imc

Let the rectangle's length be $a$ and its width be $b$. Its area is $ab$ and the perimeter is $2a+2b$. Then $A+P=ab+2a+2b$. Factoring, we have $(a+2)(b+2)-4$. The only one of the answer choices that cannot be expressed in this form is $102$, as $102+4$ is twice a prime. There would then be no way to express $106$ as $(a+2)(b+2)$, keeping $a$ and $b$ as positive integers. Our answer is then $\boxed{102}$. Note: The original problem only stated that $A$ and $P$ were positive integers, not the side lengths themselves. This rendered the problem unsolvable, and so the AMC awarded everyone 6 points on this problem. This wiki has the corrected version of the problem so that the 2015 AMC 10A test can be used for practice.