Every Information Is Important

Hi everyone,

Let $S$ and $P$ be the sum and product of $x, y$ respectively.

The first claim of Brina means $P$ can't be factorized as $k\times p$ where both $k$ and $p$ are prime numbers OR $p$ is a prime which is greater than $50$

The first claim of Brian implies $S$ can't be expressed in terms of any pair of primes AND $S<53+2$ ( $53$ is the first prime which comes after $50$ and $2$ here is because we are given that $1<x<y$ ).

Therefore, $T=\left\{ 11;17;23;27;29;35;37;41;47;51;53 \right\}$ is the set of all possible values of $S$ (honestly, I worked it out by running a simple code). Although $6$ satisfies the conditions too, it will eventually make redundant because either Brina and Brian would be able to end the game if they had it. Now, we notice that $S$ is odd i.e. one of $x,y$ is odd and the other is even, thus $P={ 2 }^{ n }\times k$ where $n\le 5$ , $k$ is odd.

The second claim of Brina guarantees there is only one way to factorize $P$ such that the sum we get is in $T$ (1)

The last claim makes it clear that there is only one way to express $S$ such that the product we get satisfies (1) . We are going to pull that $S$ out. To make such tough work easier, it would be useful to consider a Brina pair $(a,b)$ which satisfies ( $a={ 2 }^{ n } \quad AND\quad b$ is a prime) OR $a=32$ .

Brina can determine the numbers "faster" if she has the product of a Brina pair. Every $t$ in $T$ can be expressed in terms of such pairs. If there are more than one way to express it then such $t$ isn't $S$ . For example, $53=32+21=16+37$ so it's safe for us to dispose of $53$ .

$51=32+19=8+43$ fails the test too.

$47=32+15=16+31$ fails again.

$37,35,27,23,11$ suffer the same fate. Next, we deal with $17,29,41$ .

$41=32+9=16+25$ , both $32 \times 9, 16 \times 25$ are easy to determine for Brina so Brian will get confused. The same thing happens to $29=16+13=4+25$ .

Everything is done here, $17$ with its unique Brina pair $(4,13)$ is the answer.

My solution may not be the best one not creative either but I think it may be the only one, if someone has some other better solution please tell me and correct me please.

What I did was that I wrote numbers $1$ to $100$ , didn't took much time , just few minutes and then I circled every prime number including $1$ .

Now, clearly as Brian replied to Brina I know that , it tells us that whatever $x+y$ is, for every two numbers whose sum is $x+y$ must not be both prime.

And as Brina said, I can determine the numbers now it tells us that whatever $xy$ is, of all the numbers $x$ and $y$ , there would be only one case when the numbers satisfy the condition of Brian,, for every two numbers whose sum is $x+y$ must not be both prime.

Now, take every number from $1$ and try to get sum as that numbers by any two primes, if the number fails the test cut it out, go to next number until you have reached one and then try the other test on it, Brina test.

You will get $17$ as the answer.