I'm aware of the general sum and product puzzle. I got one in a math class, and am wondering if it even has solutions?
The teacher gives student A the sum $$s = x + y$$, $B$ the product $$p = xy$$. A says "I know you don't know the sum". B says "I didn't but now I know s = 136". Find $x$ and $y$.
This seems to be missing something along the lines of "x isn't equal to 1" or "s < 100", thus seems unsolvable. Does this have a solution?
If B gets a prime number $p$, then he would know that it could only be written as the product $1*p$, and would know the sum is $s=1+p$.
Every composite number $r\cdot t$ can also be written as $1$ times $rt$. So if B gets such a composite number, he cannot know the whether the sum is $r+t$ or $1+rt$. (These two values are never equal if $r>1$ and $t>1$.)
A says he knows for sure that B won't know the number. This means that A is certain that B could not have received a prime number. Most ways of splitting his sum into $x+y$ leads to $p$ being a composite number. The only way it might lead to a prime is if $(x,y)=(1,s-1)$. Therefore A can only make his claim if $s-1$ is not a prime.
So we now know that $p$ and $s-1$ are both composite numbers.
B can make this same deduction, so from A's claim he also deduces that $s-1$ is composite. The factorisation as $(x,y)=(1,p)$ leads to a composite value of $s-1$, so for B to be able to make his claim, all the other possible $(x,y)$ pairs with $x\cdot y=p$ must give $x+y-1$ a prime value, and furthermore, from this he deduces that $(x,y)=(1,p)$.
Since he claims that $x+y=136$, we must have $(x,y) = (1,135)$ and $p=135$.
We should check that B could indeed deduce this from the value $p=135$ he was given.
$135 = 1*135 = 3*45 = 9*15 = 27*5$
This gives possible values of $s-1$ of $135$, $47$, $23$, $31$, and indeed only the first is composite.