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Assume that $X$ and $Y$ are positive integers with $1<X<Y$. Mr. S knows the value $X+Y$, and Mr. P knows the value $XY$. They then have the following conversation. Mr. S says to Mr. P : "You do not know the value of $X$ and $Y$" Mr. P responds to Mr. S: "Now that you said this to me, I know the values of $X$ and $Y$." Mr. S then responds : "So do I". Find $X$ and $Y$.

My (very inadequate!) thoughts on this puzzle so far are as follows. It is clear that $X$ and $Y$ cannot be two distinct primes-else P would know their values instantly. I also thought (assuming Goldbach's conjecture (is this legit?!) that $X+Y$ cannot be even. Because if $X+Y$ were even, Mr. S could not rule out the possibility that $XY$ is the product of two distinct primes. Thus, all even numbers in $[1,Y]$ can be ruled out. The same reasoning also rules out odd positive integers that are of the form $2$ added to an odd prime, and squares of primes in $[1,Y]$. The problem as stated did not give a numerical bound on $Y$. If it had-like (say) $Y=100$, I would be tempted to try cases and see which integers can be ruled in-many seem to be ruled out, and there could well be more than one solution. Is there a general argument that works for any positive integer $Y$ that I am missing?

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    $\begingroup$ Your idea re Goldbach etc. was extremely useful in clarifying my thoughts about the numbers. $\endgroup$
    – user502266
    Jan 12, 2020 at 11:57
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    $\begingroup$ Perhaps Mr. S and Mr. P are bluffing? $\endgroup$
    – Servaes
    Jan 12, 2020 at 15:23
  • $\begingroup$ Servaes. They could be! I sincerely hope not! They wasted a lot of my time if they did. $\endgroup$
    – student
    Jan 12, 2020 at 22:52

1 Answer 1

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A solution to this problem is $X=4,Y=13$.

Proof this works

S knows that $(X,Y)$ is one of $(2,15)(3,14),(4,13),(5,12),(6,11),(7,10),(8,9)$. The product $XY$ is then one of $30,42,52,60,66,70,72$. Each of these has at least two permissible expressions as a product and so P does not initially know the two values.

P knows that $(X,Y)$ is one of $(2,26),(4,13)$. However, if $(X,Y)=(2,26)$, then S could not exclude the possibility $(X,Y)=(5,23)$ when P would initially know the two values. So, once S has spoken, P knows that $(X,Y)=(4,13)$.

Now back to S who knows that P has been able to find the values. For each possible $XY$ other than $52$ he can see that P would not be able to distinguish between the solution with $X+Y=17$ and :-

$XY=30$. $X=5,Y=6.$

$XY=42$. $X=2,Y=21.$

$XY=60$. $X=3,Y=20.$

$XY=66$. $X=2,Y=33.$

$XY=70$. $X=2,Y=35.$

$XY=72$. $X=3,Y=24.$

Proving that there are no other solutions looks extremely messy!

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  • $\begingroup$ Thank you Mr. Dolan! I suppose one could try different numerical values of $Y$ to see if more solutions emerge. It may be reasonable to ask (if other solutions exist) what proportion of $Y^2$ makes up the solution set as $Y^2\rightarrow{\infty}$. $\endgroup$
    – student
    Jan 12, 2020 at 23:10
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    $\begingroup$ You could try but the problem is that even with this small example the number of possibilities is becoming very large. My impression as I worked through possibilities is that there are likely to be very few or no others. $\endgroup$
    – user502266
    Jan 12, 2020 at 23:19
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    $\begingroup$ Remember that P starts with the knowledge that $XY=52$. $\endgroup$
    – user502266
    Jan 12, 2020 at 23:34

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