Sum Things are Products of Twisted Minds

Two people, named S and P, are talking about two numbers x and y, with x and y both integers greater than or equal to 2. S knows their sum (x+y), while P knows their product (xy); however, initially NEITHER knows x and y.

S: Hey P! I don’t know what the numbers are.
P: I’m not surprised. I KNEW you didn’t know. However, I too don’t know.
S: You don’t? Really! Then I know what the numbers are!

What are the two numbers?

Communicated by Kathryn Sievers.

Posted in General, Hard, Logic

44 Comments

  1. Steven Miller on June 14, 2012 at 2:58 am

    ayush: correct (sjm1 AT williams.edu)



  2. Steven Miller on May 27, 2012 at 1:17 am

    No. email to you bounces, email me at sjm1 AT williams.edu



  3. Maria on May 27, 2012 at 12:10 am

    is it 6 and 4



  4. mutui online on April 27, 2012 at 10:30 pm

    Right! I agree.



  5. Steven Miller on April 18, 2012 at 1:00 am

    no, there’s enough info here — email me at sjm1 AT williams.edu for a hint



  6. tqlao on April 17, 2012 at 9:46 pm

    seems like something is missing



  7. Steven Miller on March 30, 2012 at 9:26 pm

    pablo: correct



  8. Pablo Tejada on March 30, 2012 at 8:35 pm

    3 & 4



  9. Steven Miller on March 28, 2012 at 5:02 pm

    Not quite. As the numbers have to be at least two, if this is the answer then both Sum and Product know the numbers from knowing the sum or the product. If x, y >= 2, the only way x*y = 4 is if x=y=2; similarly the only way x+y=4 is if x=y=2.



  10. Anonymous on March 28, 2012 at 4:09 pm

    2 & 2



  11. Steven Miller on March 24, 2012 at 1:44 pm

    Sam: you have the right answer, but not sure if you have the full details for the solution. Email me at sjm1 AT williams.edu if you want to chat more. //s



  12. Steven Miller on March 17, 2012 at 1:51 am

    close, but not quite — email [email protected] if you want to chat. ./s



  13. Anonymous on March 16, 2012 at 9:26 pm

    4, 4?



  14. Steven Miller on February 4, 2012 at 1:56 am

    it’s not 4 and 4 — email me at [email protected] if you want a hint



  15. awesome on February 3, 2012 at 11:48 pm

    4 and 4
    because P saying he didn’t know means the product number has to have more than 4 factors…for example 2×2=4, but can only be 2 times 2. 3×3=9. 4×4=16, but it can be 4×4 OR 2×8. So if S knows the sum of the numbers then S will know…
    but can’t it work with other numbers as well?



  16. Steven Miller on January 30, 2012 at 3:52 am

    no, not 5 and 2



  17. kevin K on January 30, 2012 at 12:26 am

    5 and 2.



  18. Steven Miller on January 3, 2012 at 5:56 am

    yes //s



  19. Maninae on January 3, 2012 at 5:35 am

    Do S and P know that x and y have to be greater than or equal to 2 initially? Thanks



  20. Steven Miller on December 30, 2011 at 9:39 pm

    Elias: email me at [email protected] and I’ll send my soln so you can compare.



  21. Steven Miller on December 30, 2011 at 9:32 pm

    Elias: you’re right — the question is can you prove nothing else works?



  22. Steven Miller on December 9, 2011 at 2:04 am

    Kareem: right answer, but you didn’t provide enough logic to rule out all possibilities. Email me at [email protected] if you want to chat further about this problem.



  23. Steven Miller on November 30, 2011 at 1:38 pm

    To: derp 87.139.234.89 Submitted on 2011/11/30 at 10:39 am: you’re on the right track. your pair IS the solution — can you prove there is no other soln?



  24. Steven Miller on November 24, 2011 at 2:00 pm

    It’s not 2 and 9. why is (3,8) not a candidate for sum? email me at [email protected] for a soln / hints / discussion



  25. SAK on November 24, 2011 at 12:41 pm

    I strongly feel that the numbers are 2 and 9. Sum is 11 and product is 18.
    S has two candidate pairs (2,9) and (5,6).
    P has two candidate pairs (2,9) and (3,6).

    The common pair is 2,9.

    I imagine that I have to put P in S’s shoes and ask question like “If I was S, how would I think?”.

    However, I can’t deduce a logic which is strong enough. I will appreciate if you can either provide me the logic if my answer is correct, or hints if it is wrong.



  26. Steven Miller on November 3, 2011 at 1:48 am

    no — initially no one knows x or y; one person knows x+y a nd the other knows xy.



  27. Anonymous on November 2, 2011 at 7:47 pm

    question…the riddle says “…initially, neither knows x and y.” Does one of them know x OR y?



  28. Steven Miller on October 30, 2011 at 12:29 pm

    it is, but the answer has two distinct numbers



  29. jeff on October 30, 2011 at 5:28 am

    Is x=y allowable?



  30. Steven Miller on October 21, 2011 at 12:41 pm

    no — remember the numbers have to be integers at least 2, possibly more.



  31. William Fox on October 21, 2011 at 5:38 am

    0 and 0?



  32. Steven Miller on October 19, 2011 at 1:10 pm

    Nope.



  33. Aldo on October 19, 2011 at 11:46 am

    2 and 8



  34. Steven Miller on October 2, 2011 at 12:56 pm

    close but not quite



  35. Harsh on October 1, 2011 at 12:58 pm

    6 and 3?



  36. Steven Miller on September 26, 2011 at 1:32 pm

    You’re close. It is two small numbers, but not 2 and 6.



  37. quess on September 26, 2011 at 6:27 am

    i would say it is:

    2 and 6
    cause if P cant say numbers there have to 2 diffenrent ways to get the product
    in this case product will be 12, you can get it with 2 and 6, or 3 and 4.
    So P cant say numbers.

    But if u add those numbers u get 8 and 7.
    So S can say numbers cause there is only one ansrew if u add but two if u mulitply.

    is this correct?



  38. mahesh on September 2, 2011 at 5:38 am

    i post the answer in 8.35 am.I am really impressed by your questions.I will email you.



  39. Steven Miller on September 2, 2011 at 2:09 am

    Not quite. If you want a hint, email me at sjm1 AT williams.edu

    To whomever posted just two numbers at 8.35am: correct.



  40. Your name on September 1, 2011 at 8:03 am

    My idea holds only if x is not equal to y.N let me suppose x=2 n y=6.S knows that sum(x+y)=8 i.e. (2+6,3+5,4+4) but as i assumed x shouldnt b equal to y,(4+4) shouldnt b considered.Also P Knows that product(x*y)=12 i.e. (2*6,3*4).Since P doesnt know that actual number,then it must b x=2 n y=6..If we take x=3 and y=5,P also must have known it..I hope i m right..



  41. Steven Miller on August 21, 2011 at 3:02 am

    not quite.



  42. K on August 21, 2011 at 1:34 am

    Could the numbers be any 2 factors that are the only factors that would make up the product other than 1 and the number? For example, the number 8 has factors: 1, 2, 4, 8. Since x and y are two numbers greater than or equal to 2, then it can’t be 1 and 8, so it must be 2 and 4. That’s how P knows it’s 2 and 4, but S doesn’t: the sum of 2 + 4 = 6, and that could be x and y could either be 2 and 4, or 3 and 3. Another solution to x and y could be x = 2, y = 5. Another solution could be x = 3 and y = 3.
    Is this correct?



  43. Steven Miller on July 1, 2011 at 1:19 pm

    Try looking at special cases. Look at how the product factors. If your product is a prime number, what would this mean? Keep thinking along these lines.



  44. kc on July 1, 2011 at 11:16 am

    hw is it possible??? i hab no clue…



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