## Sum and Product

Posted by Chris on July 9, 2010 – 8:04 pm

Two perfect logicians S and P are told that integers x and y are such that:

1 < x < y and that x + y < 100.

S and P are then given the values x+y and x*y respectively and privately. S and P know they have each been given the sum and product respectively. They then have the following conversation:

P: I cannot determine the two numbers.

S: I knew that.

P: Now I can determine them.

S: So can I.

Given that they spoke the truth, what are the two numbers?

July 9th, 2010 at 8:08 pm

I messed this one up when I posted it the first time round. I have no idea where I found it. The rotten cheaters there overused a computer to crack it.

I’m sure that there are a few visitors here who can do better than that.

Cam. I found the sheep problem as well. I’ll post it soon.

July 9th, 2010 at 9:02 pm

Chris…do you have a secret access link to the old Trick of Mind site or something.if so,please share it with me.as per this question,i’ll only guess that S and P both forgot that they are the same person.

take it from me,it happens sometimes.

July 9th, 2010 at 9:06 pm

Hi Knightmare and Pequod, I simply log on at http://www.blogger.com with my email/password. But no cheating, I’ll notice if either of you suddenly turn into a genius

July 9th, 2010 at 9:21 pm

Since P cannot determine what the numbers are, there is more then one way to factor the product, which implies that at least one of the numbers is not prime. Because S could not figure out the numbers until P could, the sum could not be the sum of two primes either. By Goldbach’s conjecture, all even numbers >= 4 can be written as the sum of two primes, so the sum has to be an odd number. For odd numbers that can be written as a sum of two primes, one of them must be 2. from the answers given, no solution can be a prime plus a power of two with multiple possible arrangements. Examining pairs where these conditions are met, only 4 and 13 will hold for the answers between S and P.

4 and 13 are then x and y respectively

July 10th, 2010 at 4:24 am

Hi Nathan. You have the correct answer. I also like your reasoning very much. Sadly, I’ve not awake enough and I don’t have enough time to chek that there aren’t any flaws in it. e.g. Goldbach doesn’t guarantee unique pairs of primes.

You have also caught me off guard, because the source site definitely diidn’t invoke Godbach and the reasoning there was much more long winded.

I’m pretty sure that Cam and slavy will check your response.

PS I don’t think the source site put the constraints on x and y.

July 10th, 2010 at 4:38 am

It has already been checked Well, this is a very famous problem so there are many sources/hints/solutions available. I don’t like the usage of a conjecture in a mathematical proof, but since we are working with small numbers and the Goldbach result is still open it is definitely true for the even 2-digit numbers (actually computers have checked the validity of the conjecture up to 10^15 and more). The other thing I don’t like is that the actual computations are hidden. The Goldbach conjecture itself is not the only trick that can be applied here and if we purely compute (as suggested) from this moment on we are in a big trouble. So, I personally, prefer to see more details in the “examining pairs, where this conditions are met” direction…

July 10th, 2010 at 5:23 am

Hi all. I’m quite happy that Goldbach was used as the numbers were small (and because Goldbach is probably true). My only worry is that e.g. 20 = 13+7 = 17+3, so the pairs aren’t unique.

Slavy, obviously, I really agree with your reservations about using Goldbach. But Nathan really should have explicitly stated that it was valid as the numbers were way below 10^18 (or whatever).

I’m taking on trust that Nathan really didn’t do anything sneaky (out of sight), especially as the non-Goldbach method can get long-winded. And yes, it’s a pity that the full calculations weren’t posted.

Sadly, the other method gave a great excuse to make all sorts of other observations e.g. if the product was a prime^3 or a prime^4 the factors would be unique (in the context of the problem). I hope I’ve got that right.