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My fellow mathematicians Paul and Sam are very fond of mathematical puzzles. One day, I came up with the idea on mathematical puzzles while three of us having lunch together.

I: Hey, guys. I am thinking of two integers which are greater than or equal to 2. I will tell you the product and the sum of two numbers. Why don't you guess what two numbers are?

P: If you tell us the product and the sum, finding two integers is a piece of cake.

I: I will tell Paul only the product and Sam only the sum.

I whispered the product to Paul and the sum to Sam.

I: First, I will ask Paul. Could you figure out two numbers?

P: (after thinking) I can't. I don't know what two integers are.

I: (to Sam) How about you Sam? Do you know what two unknown numbers are?

S: I do not know either.

Surprisingly, right after Sam answered that he does not know, Paul said;

P: Now, I think I know these two numbers.

S: I also found out two numbers.

Now, a question to the reader. What are two integers?

Remark: The answer is NOT 4 and 13.

From this riddle, after answering "I don't know" twice, the answer " I know" came one after another.

If "I don't know" comes three times and then answers "I know", what would two integers be?

What is the largest possible number of "I don't know" before one can answer "I know" when solving two integers?

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  • $\begingroup$ Why do you specify that (4,13) is not the answer? $\endgroup$
    – frodoskywalker
    Commented Jan 25, 2015 at 17:51
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    $\begingroup$ (4,13) is the solution of the (most?) famous "I don't know" puzzle. Whenever I said my puzzle, someone said that the answer of the famous puzzle was 4 and 13. :-) $\endgroup$
    – P.-S. Park
    Commented Jan 26, 2015 at 12:15
  • $\begingroup$ It's not 4 and 13 because the numbers aren't specified to be within 100. $\endgroup$
    – user88
    Commented Aug 4, 2015 at 5:56

2 Answers 2

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The numbers are

6 and 2

Explanation

P gets 12 and S gets 8
P knows the numbers are (4, 3) or (6, 2), so replies I don't know
S knows the numbers are (2, 6) or (4, 4) but can not be (3, 5) as P doesn't know the answer, so still replies I don't know
Suddenly P realizes that numbers can not be (4, 3) otherwise S would have ruled out (5, 2) and guessed (4, 3) and hence (6, 2) is the answer so he replies I know the answer
S also realizes that the numbers can not be (4, 4) otherwise P could not know the number so he concludes the numbers are (6, 2) and replies I know the answer

I assume that both P and S are perfect logicians.

Edit:
Answers to the bonus questions are:

(4,4) for sequence no, no, no, yes
There may be infinite rounds of NO

I will add the details once I put down my paper work in text.

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  • $\begingroup$ Right! How about the extra questions? $\endgroup$
    – P.-S. Park
    Commented Jan 25, 2015 at 12:39
  • $\begingroup$ @P.-S.Park I missed out the extra question after I started to solve main question. I would try to solve and post my analysis here. $\endgroup$
    – Mohit Jain
    Commented Jan 27, 2015 at 4:53
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    $\begingroup$ @BmyGuest Pair (5, 2) is correct and as intended. P thinks if the number were (4, 3), S would have got 7. Now 7 can be formed only as (4, 3) or (5, 2) in given constrains ignoring order. S thinks if (5, 2) was correct answer, P could have easily guessed it as both the numbers are prime. $\endgroup$
    – Mohit Jain
    Commented Jan 27, 2015 at 4:57
  • $\begingroup$ The number of "No" is limited. If "No" is more than the limit, P and S never know two numbers. $\endgroup$
    – P.-S. Park
    Commented Jan 27, 2015 at 9:47
  • $\begingroup$ In the first place HOW did u know that P got 12 and S got 8? Just to clarify further your explanation, please. $\endgroup$
    – user14456
    Commented Aug 7, 2015 at 9:24
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In the accepted answer the second bonus question is answered incorrectly:

What is the largest possible number of "I don't know" before one can answer "I know" when solving two integers?

It should be like that:

3 don't knows: 4,4
4 don't knows: 2,8
5 don't knows: 4,6
6 and more: never

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  • $\begingroup$ Right answers! Good job. $\endgroup$
    – P.-S. Park
    Commented Aug 1, 2017 at 11:32

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