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Hardest Math Problem Ever?

  1. Jul 9, 2009 #1
    Hi, my college professor gave me this brain teaser that I've been working at for a week but to no avail. Hopefully someone here can give me some assistance, because my professor won't.

    Each natural number can be put into one of two categories: correct and incorrect. Each number will always be either incorrect or correct and will never change from one to the other.

    Here are the examples he gave us: (note that the list of numbers is never-ending, he just gave us a small list)

    Correct: 1-10, 12, 18, 20, 21, 24, 27, 30
    Incorrect: 11, 13-17, 19, 22, 23, 25, 26, 28, 29

    Over the past week, I've gotten to tell me at least this: there is no pattern, no sequence. Each number can be looked at individually and can be categorized within a few seconds, though longer numbers take a little longer to figure out (i.e. 4135). No calculator is necessary to do the math.
    ... To categorize each number, one needs to perform TWO math operations, that is +, -, x, or division. No complex math or big numbers involved. A sixth grader has the math knowledge to solve this. The result you get from doing these two operations will give you a number that has a certain number property. The property is what tells you which category it gets put into.

    Good luck!
  2. jcsd
  3. Jul 10, 2009 #2
    I think I can see it!

    A hint (if I'm correct, and none of the examples have failed):
    You don't have to use any other, unrelated number on it (eg, you just don't (without a reason that relates this number to the original nubmer) divide by two). Just look at the number itself.

    Another hint:
    Any two digit multiple of 10 will be correct.

    And the answer (if I'm correct):
    A number is "correct" if the number is divisible by the sum of its digits. Else it is incorrect.
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