Two numbers and two logicians

jason_t · #1 07-23-2005, 07:14 PM

Two perfect logicians, Godel and Tarski, are told that two integers m and n are chosen so that 1 < m < n and m + n < 100. Godel is told the value of m + n and Tarski is told the value of mn. The following dialogue takes place.

Tarski: I cannot determine the two numbers.
Godel: I already knew that.
Tarski: Now I can determine them.
Godel: As can I.

The above statements are true. What are the two numbers?

Please post solutions in white.

The Dude · #2 07-23-2005, 07:20 PM

Anaheim is not Los Angeles.

random · #3 07-23-2005, 07:24 PM

I gave up and cheated. Yeah, there was no way I was going to get that.

Piz0wn0reD!!!!!! · #4 07-23-2005, 08:00 PM

3 and 17?

jason_t · #5 07-23-2005, 08:34 PM

[ QUOTE ]
 3 and 17? 

[/ QUOTE ]

No.

Xelent · #6 07-23-2005, 09:15 PM

I just don't understand how the fact that Godel knows Tarski cannot determine the numbers would lead to them both knowing because there are so many combinations that work.

Xelent · #7 07-23-2005, 09:25 PM

nevermind I understand, I overlooked the fact that they are perfect logicians. ick, didn't realize how much work involved in this problem. I'll do it over lunch.

kurosh · #8 07-23-2005, 09:37 PM

I'm sick at the moment so I can't think very well. Once I'm better I'll be able to figure out a formula, but basically the way it's done is like this. For the m + n guy, the only requirement is that the number be greater than 7 for him not to be able to determine it. For the other guy, the number has to <=2450, >=6 and have more than two factors, not including 1 (i think that's the right term).

I'm sure there's a formula way of doing it and like I said, once I'm not feverish I'll figure it out but it'd go something in the manner of, ok, addition guy has 7. That means possible m and n values are 4/3 and 5/2. That means multiplication guy can have 10 or 12. It can't be 10 because it only has two factors besides 1 and the guy could figure it out on his own. Can it be 12? Factor combinations for 12 are 6/2 and 4/3... and I just lost it. I'll be back when I don't have a fever and can think well.

Piz0wn0reD!!!!!! · #9 07-23-2005, 09:38 PM

3 and 7!?!?!

Xelent · #10 07-23-2005, 09:59 PM

multiplying all valid products, the only standalone sum was 17. All other eligible sums had multiple combinations. My answer is 4 and 13.