A Holiday Puzzle

Discussion in 'Discussions' started by Haldurson, Dec 24, 2011.

  1. blob

    blob Member

    That north pole thing totally lost me.
     
  2. Haldurson

    Haldurson Member

    It's something you have to visualize. You may want to use a round object like a ball, or even better, a globe, as a visual aid to help. You can think of it as a non-euclidean geometry problem (geometry on the surface of a sphere). If you try to do it on a typical map, it just won't look right. Does that help?
     
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  3. blob

    blob Member

    Ok got it.
     
  4. Godwin

    Godwin Member

    Awww no more riddles?
     
  5. Haldurson

    Haldurson Member

    Someone else come up with some then. These are all from memory, which is failing me right now. If I think of any more I'll be sure to post them, but these are probably my best, or at least, my favorites.
     
  6. Haldurson

    Haldurson Member

    OK, I did remember one -- not exactly a riddle, more of a calculus proof that I read years ago in "The Journal of Irreproducible Results" (kind of a cult magazine among science and engineering students)

    Here's how it goes -- besides the obvious end result, tell me exactly what is wrong with the following proof:

    1. X^2 = X*X
    2. X^2 = X+X+X+X... (X times)
    3. Derivative(X^2) = Derivative( X+X+X+X... (X times) )
    4. 2*X = 1+1+1+... (X times)
    5. 2*X = X
    6. 2 = 1
     
  7. lccorp2

    lccorp2 Member

    Let's see if I still remember this:

    First off, most obvious--2 is false if x is negative. -2^2 = 4. No matter how many times you sum -2, you will never get a positive number, and it is meaningless to say that a series has -2 terms anyways.

    Second point is also between steps two and three--the number of terms and value of said terms are linked, which needs to be accounted for (I.E, x times is NOT independent of x). In other words, we can do:

    d/dx[3x] = d/dx[x + x +x] = d/dx[x] + d/dx[x] + d/dx[x] = 1 + 1 + 1 = 3 as 3 is a constant w.r.t x.

    HOWEVER,

    d/dx[ x + x + x + ... (x times) ] =/= d/dx[x] + d/dx[x] + d/dx[x] ... (x times) because x is a variable.

    Final point is that the derivative d/dx[x^2] = 2x is defined along a continuous function. In part 2, we are taking X on the right hand side as discrete points. Since the left hand side of the equation is a derivative defined over a continuous function and the right hand side is defined over discrete points, which are defined only for positive integers to boot.

    Hence, by these three proofs, the equations in the problem statement are erronous.

    Here's another 2=1 puzzle, but much easier than Haldurson's:

    1. a = b
    2. a+a = a+b
    3. 2a = a+b
    4. 2a-2b = a+b-2b
    5. 2(a-b) = a-b
    6. 2 = 1

    Problem lies in which step?
     
  8. Haldurson

    Haldurson Member

    I'm not sure if I've seen this exact problem before, but I've seen similar, so I knew what the problem was going to be before I actually read the whole thing:

    You can't go from 5 to 6 because you already established that a=b, hence a-b = 0, and you cannot divide by zero.
     
  9. Haldurson

    Haldurson Member

    This one should be easy -- It was in the copy of "Scientific American Mind" that is on sale right now, so my memory of it is fresh:

    The day before yesterday I was 29 and next year I will be 32. What is the only day of the year when this can be true?
     
  10. Godwin

    Godwin Member

    @Haldurson: your '2=1' problem reminds me of another, it was actually something I came up with together with a friend in high school and which we presented to our math teacher:

    i is the complex number

    i = (-1)^1/2 (square root of -1)
    so i^2=-1
    so [(-1)^1/2.(-1)^1/2]^2 = -1
    so [(-1).(-1)]^(2.1) = -1
    so [1]^2 = -1
    so 1 = -1

    I think the answer to your age one is that the person's birthday is 31st of december and it's the first of january.
    the 30th you're 29, the 31st you're 30, so in the new year that starts the first you'll be 31, so next year you'll be 32 (at the very end of the year).

    EDITED to correct the silly mistake that doesn't alter the problem.
     
  11. Haldurson

    Haldurson Member

    Your solution is correct.

    Good one :cool:.
    I think the mistake is that it should actually be:
    [(-1)^(1/2)*(-1)^(1/2)]^2 = [(-1)*(-1)^(2*1/4)]=-1
     
  12. Godwin

    Godwin Member

    Nope. It's hard to write math formulas like this, but the square root of 2 times the square root of two is two, because the power of 1/2 plus the power of another half equals the power of one.

    But I did make a mistake :eek:, silly me!

    It should have been [(-1).(-1)]^(2.1) (. = x = * = multiplication). Powers add up when the numbers that are 'powered' are multiplied and the same (x^y*x^z = x^(y+z)).

    But in the end, it wouldn't even matter: (-1)^x*(-1)^x = [(-1)*(-1)] ^x = [1]^x = 1
     
  13. blob

    blob Member

    *crawls into a foetal position and cries silently*
     
  14. Godwin

    Godwin Member

    Lol why?
     
  15. blob

    blob Member

    Math's always been beyond me, and you two seem to enjoy so much putting billions of x, y -1 and ^something everywhere :p
    It reminds me why I could never really get into coding ! Hehehe
     
  16. Haldurson

    Haldurson Member

    I used to be much better at math when I was younger than I am now -- plus my job didn't give me much opportunity to use very much math beyond algebra, except on rare occasions (I actually needed calculus like 20 years ago for one project -- that's how rare it is that I can remember it). I know I've forgotten a lot (though I'm still able to help my niece with her math homework). Granted she's just started High School, and not even doing pre-calculus yet, but still.

    Beyond that, I've had some health-related problems lately that have interfered with my math ability. I was a Senior Systems Analyst at a moderately sized international company, but had some health problems and have been on disability. I also strongly suspect (since it runs in my family) that I have symptoms of early onset Alzheimers. But since I still am pretty above-average with math (and keep my mind active with stuff like this), my best friend/ex-boss thinks I'm being crazy. But I do remember how much easier some things used to come to me. I just don't have the focus and concentration for it that I used to have.
     
  17. blob

    blob Member

    Aging and alzheimer both definetly suck but keeping your brain active is probably the best way to deal with it. If I cant deal with maths now im probably gonna turn into a vegetable by that time :)
    Thankfully I have a niece too, maybe she ll be a good excuse to get back into maths... Well. Probably not though.
     
  18. Haldurson

    Haldurson Member

    Research definitely shows that although the disease itself is not affected by keeping an active brain, the symptoms are. I know my dad started doing crossword puzzles after he was diagnosed. When he was younger he was pretty good at chess, but had given it up (no joke -- after many games, I finally beat him once, and then he never played again). I tried to get him to play again after he was diagnosed, but he really couldn't play anymore. I feel the same way now -- a friend I used to beat all the time, I now can't touch.
     
  19. Godwin

    Godwin Member

    Well it's logical that the things you invest only a minority of your time and energy in are the things you forget or confuse the first. So I'd make sure that the most important things deserve the most attention. Like happiness. Family. After that I'm sure crosswords or maths may help you stay sharp.
    If you're willing to accept that things change and you lose abilities you had without that making you unhappy, you're in for a happy ride :)
     
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