A conversation at the Brilliant water cooler

First, since both $x$ and $y$ exceed $1$ their product $P$ cannot be prime. Now if both $x$ and $y$ were prime then Calvin would know what the two numbers were, (and hence also their sum), as a result of knowing $P$ . For example, if $P = 10$ then the two numbers could only be $2$ and $5$ , giving $S = 7$ . Thus at least one of $x, y$ is a composite number, and so the prime decomposition of $P$ must meet one of the following forms:

(i) $p_{1}^{a_{1}}*p_{2}^{a_{2}}*...*p_{k}^{a_{k}}$ for primes $p_{n}$ with $k \ge 3$ and each $a_{j} \ge 1,$

(ii) $p_{1}^{a_{1}}*p_{2}^{a_{2}},$ where $a_{1},a_{2}$ are positive integers at least one of which exceeds $1,$ or

(iii) $p^{k}$ for $k \ge 4$ .

Now by the A.M.-G.M. inequality, we know that $(x + y) \ge 2\sqrt{xy}.$ So if $(x + y) \le 20$ , as Aaron has stated in his hint, we know that $\sqrt{xy} \le 10 \Longrightarrow xy \le 100.$

Now that, given Aaron's assurance, Calvin has enough information to determine $x$ and $y$ , we know that $P$ is such that, of the possible sums, only one is strictly less than $21$ . For example, If $P$ had been, say, $66$ , then since $66 = 2*33 = 3*22 = 6*11$ the possible sums that Calvin could have originally guessed at would have been $35, 25$ and $17$ . Once he was told that $S$ did not exceed $20$ he could then determine that $S = 17$ and thus that $x$ and $y$ were $6$ and $11$ . But from Aaron's perspective, in this case it would not be enough to know that Calvin has determined the two numbers, since, for example, $52 = 2*26 = 4*13$ would have a unique sum of $17$ under $21$ as well. So we need to find a product $P$ that has a single potential sum under $21$ that no other product $\le 100$ has as well.

Now comes the tedious work. We need to go through all the products $\le 100$ that meet one of the conditions (i), (ii) or (iii), find those that have a single potential sum under $21$ and compare. Here is a list of the candidates in the form $(X,d)$ where $X$ is the product and $d$ is the single sum strictly under $21$ :

$(44,15), (50,15), (51,20), (52,17), (54,15), (63,16), (66,17), (75,20),$

$(78,19), (80,18), (81,18), (88,19), (90,19), (96,20), (100,20).$

All the sums are duplicated except in the case $(63,16)$ . So since $16$ is the only possible unique solution given the information provided by Aaron, Calvin can conclude that $S = 16$ , and once he states that he has determined $S$ Aaron can then conclude that $P = 63$ .

Thus $S + P = 16 + 63 = \boxed{79}$ .

This solution is a visual representation of Brian’s approach. We further improve on it by drawing the matrix who rows are the sums S, columns are the product P, and the entry is colored if there are integers $x, y, > 1$ such that $x + y = S, xy = P$ .

This is an infinite table, so let’s not reproduce it as yet. Setting up this table requires a lot of tedious work (which Brian mentioned). Here are the first 10 rows, to give you an idea of what is happening

For example, $x = 2, y = 2$ gives us $S = 4, P = 4$ so the cell $(4,4)$ is marked.
Similarly, $x = 2, y = 3$ gives us $S = 5, P = 6$ so the cell $(5, 6)$ is marked.
We are not allowed to use $x = 1, y = 4$ , so the cell $(1+4, 1 \times 4)$ is not marked.

We will aim to remove rows and columns whenever we can. Once a row or column has no marked squares in it, this indicates that there are no valid solutions, and so we can delete them. EG we can remove the first 3 rows.

Now, let’s start the analysis. Calvin says “I cannot determine S at this point”.
If a number has 2 factors, of the form $1, p$ , then there is no $x, y > 1$ .
If a number has 3 factors, of the form $1, p, p^2$ , then there is no $x, y > 1$ .
If a number has 4 factors, of the form $1, p, q, pq$ , then there is only 1 set of $x, y > 1$ , and so Calvin can determine S.
If a number has 4 factors of the form $1, p, p^2, p^3$ , then there is only 1 set of $x , y > 1$ , and so Calvin can determine S.
Conversely, if a number has 5 or more factors, then there are 2 distinct ways to express the product, and hence Calvin cannot determine S. Thus, we can eliminate numbers with 4 or fewer factors, while keeping the rest of the rows. (We still have an infinite table, so I'm not reproducing it here.)

Arron says “S does not exceed 20”, which means that we can get rid of columns without any entries in the top 20 rows. This gives us

Since all Calvin needs to know is the value of S to uniquely determine these numbers, this means that we’re looking for columns with exactly 1 entry. Thus, we can delete the following yellow columns After deleting these columns, we can delete empty rows to get:

Now, Arron’s comment about “then I will know what P is” means that there is a unique number in the row that we're interested in. Thus, we can get rid of the following rows to obtain:

Hence, the only possibility is $S = 16, P = 63$ .