DeToasty3 wrote:

We start with a miraculous claim which is the pith of this problem:

Claim 1. Bob's number must be $2$ .

Proof. Before we can get our hands dirty, we shall prove a few subclaims:

Subclaim 1.1. Alice cannot have the number $1$ .

Proof. If she did, then Alice would know that Bob's number is greater than hers.

Subclaim 1.2. Alice cannot have the number $40$ .

Proof. If she did, then Alice would know that Bob's number is less than hers.

Subclaim 1.3. $1$ is a possibility for Bob's number.

Proof. If $1$ is Bob's number, then he knows that Alice's number is larger than his.

Subclaim 1.4. $2$ is a possibility for Bob's number.

Proof. If $2$ is Bob's number, then he knows that, because Alice's number cannot be $1$ , Alice's number is larger than his.

Subclaim 1.5. $39$ is a possibility for Bob's number.

Proof. If $39$ is Bob's number, then he knows that, because Alice's number cannot be $40$ , Alice's number is less than his.

Subclaim 1.6. $40$ is a possibility for Bob's number.

Proof. If $40$ is Bob's number, then he knows that Alice's number is less than his.

We have deduced that Bob's number can be either $1$ , $2$ , $39$ , or $40$ . Of these four numbers, only $2$ is prime, hence we have proved the claim. $\square$

Now that we know that Bob's number is $2$ , we may use Alice's final sentence to finish the problem. Multiplying $2$ by $100$ is merely equivalent to computing the expression $\underbrace{2 + 2 + 2 + \cdots + 2 + 2}_{100\text{ times}}$ , from which we produce the number $200$ . Now, we must search for perfect squares. Before we do this, we have to bound the range of possible values of this perfect square. To do this, we introduce another claim:

Claim 2. The perfect square must be contained in the interval $(202,240)$ .

To prove this claim, we introduce two subclaims.

Subclaim 2.1. The perfect square must be strictly greater than $202$ .

Proof. As Alice cannot have the number $1$ , she must have at least the number $2$ . However, since Bob has the number $2$ , Alice must have at least the number $3$ . Thus, the perfect square must be at least $200+3=203$ , hence proven.

Subclaim 2.2. The perfect square must be strictly less than $240$ .

Proof. As Alice cannot have the number $40$ , she must have at most the number $39$ . Thus, the perfect square must be at most $200+39=239$ , hence proven.

Combining these two subclaims, we have that the perfect square must be contained in the interval $(202,240)$ , as desired. $\square$

Now, we must search for perfect squares in the interval $(202,240)$ . Upon squaring $15$ , we find that the number obtained is $225$ , which, by sheer luck, is in the interval $(202,240)$ . However, upon squaring $14$ and $16$ , we obtain the numbers $196$ and $256$ , neither of which are in the interval. It is well-known that $f(x)=x^2$ is strictly increasing as $x$ increases from $0$ . Thus, if $14^2$ and $16^2$ are out of the interval $(202,240)$ from opposite sides of the interval, we may conclude that $15^2=225$ is the only possible perfect square. This means that Alice's number is $225-200=25$ .

Finally, we may take the sum of Alice's and Bob's numbers, which are $25$ and $2$ , respectively, for a final answer of $25+2=\boxed{\textbf{(A) }27}$ , and we are done. $\blacksquare$

Remark. This problem is clearly unsuitable for the AMC 10. Proving even one of these claims by itself proves to be a significant challenge, perhaps too much for tenth graders.

Truly a marvelous solution, the likes of which have not been seen in a century.