6
$\begingroup$

I heard from a clever person that for the problem mentioned before: I don't know the two numbers... but now I do a solution exists even if we change 100 to infinity.

So the formulation would be next:

Two perfect logicians, Summer and Proctor, are told that integers x and y have been chosen such that $x>1$ and $y>1$. Summer is given the value x+y and Proctor is given the value x⋅y. They then have the following conversation.

Proctor: "I cannot determine the two numbers."
Summer: "I knew that."
Proctor: "Now I can determine them."
Summer: "So can I."

Given that the above statements are true, what are the two numbers?

I don't know whether it is true. Can you find the solution or prove that it doesn't exists? Can you find a general solution for any integer $N$ if we have limitation $x+y<N$?

$\endgroup$
  • $\begingroup$ Sorry but wouldn't the exact same x and y work in this case? $\endgroup$ – d'alar'cop Oct 10 '14 at 12:28
  • $\begingroup$ @d'alar'cop, why? I don't see a reason for this assumption. But if you can prove it, go for it. $\endgroup$ – klm123 Oct 10 '14 at 12:34
  • $\begingroup$ It seems that by the argument in the other thread, there are actually many solutions in general. So, any of those will be a fit for this. In fact, the other answer doesn't imply that the solution provided was even the only one under 100 (and I also suspect the other one may be wrong - what about 70 (with factors 10 and 7 as counterexample)?). So, I suspect there are many solutions, maybe if the question were "what is the lowest x (or y)... etc". Not sure to be honest. $\endgroup$ – d'alar'cop Oct 10 '14 at 12:42
  • $\begingroup$ @d'alar'cop, I know for sure that for x<100,y<100 there is only one solution. $\endgroup$ – klm123 Oct 10 '14 at 13:38
  • $\begingroup$ Oh ok. So then this question is sort of like "are there any other solutions besides 17,52?", right? $\endgroup$ – d'alar'cop Oct 10 '14 at 13:43
5
$\begingroup$

The solution to the restricted version still works, but it is no longer unique. I'll give a proof that the previous solution still works. According to Joe Z.'s answer, Summer is given $x+y=17$, and Proctor is given $x \cdot y=52$. Also, to avoid confusion I'm going to call Proctor male and Summer female.

Because $x \cdot y=52$, Proctor knows the numbers could be either 2 and 26 or 4 and 13, but he has no way to determine which they are.

Summer knows that since $x+y=17$, the numbers could be any of the pairs $(2,15), (3,14),(4,13),(5,12),(6,11),(7,10),$ and $(8,9)$. None of those pairs are two prime numbers, so she knew that Proctor wouldn't be able to tell what the numbers are based on their product.

Now that Proctor knows that Summer knew he wouldn't be able to tell the numbers by their product, he can reconsider the two possibilities - if the numbers we 2 and 26, then the sum would be 28. If the Summer had the number 28 then her list of possibilities would include the pair (5,23), both of which are prime. Then she would not have known that he would not be able to tell. So he then knows that the numbers are 4 and 13.

Now that Summer knows that Proctor knows the numbers, she can go through a similar process for her list of possible numbers.

  • If the numbers were 2 and 15, then the product would be 30. Then Proctor's list of possibilities would be $(2,15), (3,10),$ and $(5,6)$. If this had been the case, then he would not have been able to eliminate $(5,6)$ because none of the pairs that sum to 11 are a pair of prime numbers. Thus Proctor would not have been able to know the numbers yet.
  • If the numbers were 3 and 14, Proctor's list would include $(2, 21)$, which could not be eliminated.
  • If the numbers were 5 and 12, Proctor would be unable to eliminate $(3, 20)$
  • For 6 and 11, Proctor would be unable to eliminate $(2, 33)$
  • For 7 and 10, Proctor would be unable to eliminate $(2, 35)$
  • For 8 and 9, Proctor would be unable to eliminate $(3,24)$

By process of elimination, Summer knows that the numbers are 3 and 14.


As you can see from my analysis, at no point did the restriction of $x+y<100$ factor in to Summer or Proctor's ability to determine the numbers. This should actually be pretty obvious because for $x,y>1$, $x+y\le x\cdot y$, so $x\cdot y<100\rightarrow x+y<100$. In other words, Summer knows the sum, and Proctor knows the product is less than 100 so he also knows that the sum is less than 100, leaving the restriction as merely a help to people trying to solve the puzzle.

The only way the restriction would affect this would be if Proctor could eliminate one or more pairs because $N$ was less than the sum of one of the pairs he would not have been able to eliminate otherwise. For example, if $N=36$, then Proctor would be able to eliminate $(2, 35)$, which would mean 7 and 10 would have still been a valid possibility based on what Summer knew.

So I think that the question of generalizing this can be split into two parts:

  • As $N$ gets larger, do more solutions become viable?
  • For smaller $N$, do different solutions become viable?

EDIT: According to this page, there are at least three more possible solutions to this problem: $x=16,y=111$, $x=201,y=556$, and $x=421,y=576$. It's under the section "Sum-Product Puzzle: (Kiltinen/Young unbounded version)".

$\endgroup$
  • $\begingroup$ That's very clear. Thanks. I made a comment to this effect but this is solid ++ $\endgroup$ – d'alar'cop Oct 10 '14 at 18:49
  • $\begingroup$ "If the numbers were 3 and 14, Proctor's list would include (2,21), which could not be eliminated."... ? $\endgroup$ – d'alar'cop Oct 10 '14 at 19:19
  • 1
    $\begingroup$ @d'alar'cop product would be 42, possibilities are (2,21),(3,14),(6,7). For both (2,21) and (3,14) Summer would know Proctor couldn't know, so Proctor still wouldn't know the numbers. I didn't want to write out the full explanation each time, but I'm following the same logic as what I wrote out for (2,15) $\endgroup$ – Rob Watts Oct 10 '14 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.