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badreferences

  • 2 years ago

Here's another brainteaser for you bored folks. "A teacher says: I'm thinking of two natural numbers greater than 1. Try to guess what they are. The first student knows their product and the other one knows their sum. First: I do not know the sum. Second: I knew that. The sum is less than 14. First: I knew that. However, now I know the numbers. Second: And so do I. What were the numbers? "

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  1. lgbasallote
    • 2 years ago
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    ahh i posted something like this before...

  2. lgbasallote
    • 2 years ago
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    though that seems harder..

  3. badreferences
    • 2 years ago
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    :P Maybe I copy-pasted from you. Ah, well I didn't. I have the answers, though. :D

  4. badreferences
    • 2 years ago
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    YES, there is enough information.

  5. apoorvk
    • 2 years ago
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    @FoolForMath posted this about a week back.

  6. badreferences
    • 2 years ago
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    Whoa, cool. :D

  7. badreferences
    • 2 years ago
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    FFM, maybe you go to the same college I do.

  8. FoolForMath
    • 2 years ago
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    This seems to be a hard problem.

  9. badreferences
    • 2 years ago
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    Or maybe you're the one assigning these difficult questions. >:|

  10. FoolForMath
    • 2 years ago
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    No, I didn't posted it, I barely post Google-able questions :P

  11. badreferences
    • 2 years ago
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    I found the answer through the wonders of Google. :P

  12. FoolForMath
    • 2 years ago
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    The numbers were 2 and 9. And here comes the entire solution. There shall be two natural numbers bigger than 1. First student knows their product and the other one knows their sum. The sum is smaller than 14 (for natural numbers bigger than 1), so the following combinations are possible: 2 2 ... NO - the first student would have known the sum as well 2 3 ... NO - the first student would have known the sum as well 2 4 ... NO - the first student would have known the sum as well 2 5 ... NO - the first student would have known the sum as well 2 6 2 7 ... NO - the first student would have known the sum as well 2 8 2 9 2 10 2 11 ... NO - the first student would have known the sum as well 3 3 ... NO - the first student would have known the sum as well 3 4 3 5 ... NO - the first student would have known the sum as well 3 6 3 7 ... NO - the first student would have known the sum as well 3 8 ... NO - the product does not have all possible sums smaller than 14 (eg. 2 + 12) 3 9 ... NO - the first student would have known the sum as well 3 10 ... NO - the product does not have all possible sums smaller than 14 4 4 4 5 4 6 ... NO - the product does not have all possible sums smaller than 14 4 7 ... NO - the product does not have all possible sums smaller than 14 4 8 ... NO - the product does not have all possible sums smaller than 14 4 9 ... NO - the product does not have all possible sums smaller than 14 5 5 ... NO - the first student would have known the sum as well 5 6 ... NO - the product does not have all possible sums smaller than 14 5 7 ... NO - the first student would have known the sum as well 5 8 ... NO - the product does not have all possible sums smaller than 14 6 6 ... NO - the product does not have all possible sums smaller than 14 6 7 ... NO - the product does not have all possible sums smaller than 14 So there are the following combinations left: 2 6 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 (it is impossible to create a pair of numbers from sum 8, so that the product would have an alternative sum bigger than 14 ... eg. if 4 and 4, then there is no sum – created from their product 16 – bigger than 14 – eg. 2 + 8 = only 10) 2 8 2 9 2 10 3 4 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 3 6 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 4 4 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 4 5 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 The second student (knowing the sum) knew, that the first student (knowing the product) does not know the sum and he thought that the first student does not know that the sum is smaller than 14. Only 3 combinations left: 2 8 ... product = 16, sum = 10 2 9 ... product = 18, sum = 11 2 10 ... product = 20, sum = 12 Let’s eliminate the sums, which can be created using a unique combination of numbers – if the sum is clear when knowing the product (this could have been done earlier, but it wouldn’t be so exciting) - because the second student knew, that his sum is not created with such a pair of numbers. And so the sum can not be 10 (because 7 and 3) – the second student knew, that the first student does not know the sum – but if the sum was 10, then the first student could have known the sum if the pair was 7 and 3. The same reasoning is used for eliminating sum 12 (because 5 and 7). So we have just one possibility – the only solution – 2 and 9. And that’s it.

  13. KatrinaKaif
    • 2 years ago
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    This is going to take ages to read.

  14. FoolForMath
    • 2 years ago
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    I agree.

  15. badreferences
    • 2 years ago
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    @FoolForMath >:| Hey, I didn't get the question and decide immediately, "Let's Google this!"

  16. lgbasallote
    • 2 years ago
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    hmm...FoolForMath Best Answer 1 No, I didn't posted it, I barely post Google-able questions :P ^is he referring to my goggle-able questions -____- haha

  17. FoolForMath
    • 2 years ago
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    @badreferences: I know, but I have seen this one before. I often read puzzles from here and there.

  18. badreferences
    • 2 years ago
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    Ahaha, I thought this one was cool. And then I Google'd it, and all the mystery dissolved. :|

  19. FoolForMath
    • 2 years ago
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    Google is a necessary evil in this case.

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