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Thread: Math Challenged

  1. #91
    This seems to be cleared. Thanks mr. obvious.

    Now who's posting the next question? o_O

  2. #92
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    Quote Originally Posted by mr obvious View Post
    So, yes, TRS's proof is complete and does not require any additional statements, even if it is not fluently stated. For example, see here:
    http://mathforum.org/library/drmath/view/55855.html
    Ha! Check it out, even the statement "or it is prime itself" can be removed from that proof!
    Quote Originally Posted by The_Radiation_Specialist View Post
    Now who's posting the next question? o_O
    Homo Bibiens, but he hasn't posted since:
    Quote Originally Posted by Homo Bibiens View Post
    I will take a little time to try to think of a good question

  3. #93
    Homo Bibiens hasn't been online since he posted that so I assume he is away or very busy...

    Moving swiftly on...

    Prove by induction

    11^n - 1 is divisible by 10 for all positive integer values of n.

  4. #94
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    Quote Originally Posted by The_Radiation_Specialist View Post
    Prove by induction

    11^n - 1 is divisible by 10 for all positive integer values of n.
    Well, it's certainly true for n=1 (or 0, or 2, or 3... )

    So, assume for some n that (11^n - 1) is divisible by 10, that is (11^n - 1) = 10A for some integer A.

    Then (11^(n+1) - 1) = 11(11^n - 1) + 10 = 11(10A) + 10 = 10(11A + 1), and since (11A+1) is also an integer, 11^(n+1)-1 is also divisible by 10.

    QED

    My question? Solve this
    Last edited by hhEb09'1; 2007-Sep-29 at 05:37 PM.

  5. #95
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    Seven mixed fruits will work, I think. I'm assuming tax/tip, etc. are included or not part of the $15.05 total.

  6. #96
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    There's at least one more answer, I'll hold out for it.

  7. #97
    Quote Originally Posted by hhEb09'1 View Post
    There's at least one more answer, I'll hold out for it.
    Please do, this is an interesting question and one which might only be solved with an exhaustive search program.

  8. #98
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    The old aphorism, many a truth is shared in jest, certainly seems appropriate in this case! The references to the knapsack and traveling salesman problems are not irrelevant.

  9. #99
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    one mixed fruit: $2.15
    two hot wings: $7.10
    one sampler: $5.80

    solving hh's puzzle: priceless!

    It's a shame that the answer isn't unique. I wonder if the answer would be unique if the fruit were $2.25 and the sampler $5.70? (And no, that's not my next question!)

  10. #100
    Cool, how did you do it?

  11. #101
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    Quote Originally Posted by pghnative View Post
    It's a shame that the answer isn't unique.
    Seems even more like a cruel joke on a waitperson, right?

    I even sent the artist an email asking about that. But surely some other geeks have attacked it.

    You and mr obvious may share the next honors (or, how about, next one of you to post a question gets to post the next question?)

    PS:
    Quote Originally Posted by pghnative View Post
    I wonder if the answer would be unique if the fruit were $2.25 and the sampler $5.70? (And no, that's not my next question!)
    one mixed fruit: $2.25
    one french fries: $2.75
    three side salads: $10.05

  12. #102
    Well the answer can be obtained by a simple Google search. I'm interested if Pghnative found it by some algorithm or some exhaustive search.


  13. #103
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    Quote Originally Posted by The_Radiation_Specialist View Post
    Well the answer can be obtained by a simple Google search.
    Ah, the author has commented in his blag (sic, another xkcd joke) that he thought the problem had one solution. "Long story short, it claimed 2.15*7 did not equal 15.05, and so it missed the second answer in the search, though it found the first just fine"

    Floating point math was the culprit

  14. #104
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    pghnative may have the honors for asking a question since I have already done so in this thread and would like some variety.

  15. #105
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    Quote Originally Posted by The_Radiation_Specialist View Post
    Well the answer can be obtained by a simple Google search. I'm interested if Pghnative found it by some algorithm or some exhaustive search.

    By a semi-exhaustive search. Used a spreadsheet to let me test permutations quickly. Since there had to be an odd number of the first 4 items, that cut down on the possible permutations.

    OK, so there's this bar where you can get a free beer if you know a secret code. You go into it, and watch for a while. Guy # 1 comes in, the bartender says "6" and the guy answers "3" and gets a free beer. Guy # 2 comes in, the bartender says "12" and the guy answers "6" and gets a free beer. Finally, guy # 3 comes in, the bartender says "14" and the guy says "8", and gets a free beer.

    The bartender looks at you and says "22" --- what do you say?

  16. #106
    9 (10 if you count the dash).

    Todd

  17. #107
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    Weirdly enough, seventh graders in the class I was teaching on Friday were talking about this function. They claimed that every number became four, if you continued iterating the function. And, of course, that's where it stops.

  18. #108
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    Quote Originally Posted by tdvance View Post
    9 (10 if you count the dash).

    Todd
    correct --- your turn

  19. #109
    144 of course

  20. #110
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    Quote Originally Posted by UFOvsUSO View Post
    144 of course
    Welcome to BAUT!

    21.

    Which, if anybody hasn't already noticed, is exactly half of 42.

  21. #111
    Ok....

    Two cars play "chicken" on a perfectly straight road, starting one mile apart. These "magical" cars accelerate instantly to 30 miles per hour and drive toward each other, ultimately crashing. Exactly halfway between the cars is a "magical" fly. This fly starts instantly at 5 miles per hour toward one of the cars. As soon as he contacts that car, he reverses course instantly with no deceleration and flies 5 miles per hour toward the other car. Upon contacting that car, it again instantly reverses direction and returns to the first car, and so on and so on. This continues until the two cars crash with the fly flattened in between them.

    How far does the fly ultimately travel, adding all the distance it flies in both directions all the time from when the cars start a mile apart to the time they collide?

    (Hint, it can be done without resorting to geometric series).

  22. #112
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    Quote Originally Posted by tdvance View Post
    Two cars play "chicken" on a perfectly straight road, starting one mile apart. These "magical" cars accelerate instantly to 30 miles per hour and drive toward each other, ultimately crashing. Exactly halfway between the cars is a "magical" fly. This fly starts instantly at 5 miles per hour toward one of the cars. As soon as he contacts that car, he reverses course instantly with no deceleration and flies 5 miles per hour toward the other car. Upon contacting that car, it again instantly reverses direction and returns to the first car, and so on and so on. This continues until the two cars crash with the fly flattened in between them.
    Shouldn't the fly's speed be greater than the car's speed? Otherwise, after it contacts the first car, the car will pull away from it and the fly will never be able to catch up to get caught between them when they crash.

    Maybe you meant, 5 mph relative to the following car? Which would be a speed of 35 mph, for the fly?

  23. #113
    oops.
    I think you need to check your numbers.

  24. #114
    The numbers may not be right but if they were, the answer would be the speed of the fly multiplied by the time it takes the cars to crash, assuming the speed of the fly is faster than the speed of the car (relative to a point on the road).

  25. #115
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    220 feet.

  26. #116
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    Which is TRS's answer with a calculator. Assuming 5MPH.

  27. #117
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    Quote Originally Posted by mike alexander View Post
    220 feet.
    That's half of what I get. The cars meet in one minute, right?

    And where will the fly be when they crash?

  28. #118
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    Quote Originally Posted by mr obvious View Post
    To DanishDynamite:

    I am not saying Euclid's proof is incorrect. I am saying your claim that TRS's proof is incomplete is incorrect. I have looked at the Wiki page you linked to. There, Euclid simply states that the product of some finite number of primes plus one is either prime or must factor a prime that wasn't in the product. He made no claims that the finite set of primes was exhaustive. Since TRS claimed (or rather strongly implied) at the very beginning of his proof that he is enumerating all primes P1, P2, etc until Pq as the largest prime, he can skip the extra step.

    So, yes, TRS's proof is complete and does not require any additional statements, even if it is not fluently stated. For example, see here:
    http://mathforum.org/library/drmath/view/55855.html

    He says there are two possibilities, that M+1 is prime, or that M+1 divides the existing, enumerated primes. Note that he does not include a random prime that divides M+1 and was not enumerated. Or see here:
    http://www.jcu.edu/math/vignettes/primes.htm

    You see, they also do not postulate the existence of some mysterious prime pw > pq. In all proofs that I see where the primes are listed exhaustively and there is a maximum designated, the only comment is that if M+1 is not prime, it must factor into an existing, enumerated prime, but it can't because there would be a remainder of one.

    Again, here:
    http://www.math.utah.edu/~pa/math/q2.html

    You see in the upper box of the proof again, they do not list the condition where M+1 can factor into some mysterious unlisted prime.

    In all of these cases, the argument is as I have outlined (although in most cases, the argument is flipped - but still equivalent). There is no need to postulate that there may be some Pw, where Pw wasn't on the list of the primes being multiplied, that divides M+1.

    Since I have provided many sources and have explained why your understanding of the Wiki section is incorrect, I hope you'll take back your unnecessarily sarcastic and rude comments. I also note that you did not rise to the challenges posed by me in the preceding posts, which were not addressed in any of your previous explanations.
    My comments were not sarcastic. I don't see any new version of Euclid's proof here, I just see a missing step in the proof. You apparently feel this is a new version and hence I encouraged you, and still encourage you, to send this in to the appropriate Mathematical body in charge of rendering verdict on these things.

    If it truely is a new version, you could get your name in the history books.

  29. #119
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    Quote Originally Posted by hhEb09'1 View Post
    That's half of what I get. The cars meet in one minute, right?

    And where will the fly be when they crash?
    Stuck on the fender of the car it meets first.

  30. #120
    Oops--I guess I did misremember the problem wrong. But yes, the solution is to find out how long (in hours) the cars take to crash, and multiply that by 5 miles per hour, to get the distance in miles the fly flew (or convert to more reasonable units!). So, it takes one minute for the cars to collide. In one minute, the fly can go 1/12 of a mile, or 440 feet. So, HeebeeJeebee99 or whatever that is, wins for two reasons--being first to point out my screw up..I mean "Lesson Learned" (at work, a manager suggested I never say "screw up" in front of budget people, but "lessons learned" instead), and for getting the correct answer to the flawed problem! (that would be the real answer, since the fly would be stuck to the one bumper flipping back and forth an infinite number of times per second till the crash).

    Todd

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