Page 2 of 27 FirstFirst 123412 ... LastLast
Results 31 to 60 of 783

Thread: Math Challenged

  1. #31
    Join Date
    Sep 2007
    Location
    Lost among the 160,000+ members
    Posts
    8,002
    Quote Originally Posted by DanishDynamite View Post
    Very good. But is this the only possible answer? (Hint: No)
    8x6
    12x5

    l assume you do not consider 6x8 and 5x12 to be distinct from the above.

    Do you also allow 0x0?

  2. #32
    Join Date
    Apr 2007
    Posts
    762
    Quote Originally Posted by DanishDynamite View Post
    A fisherman catches a number of fish. The 3 heaviest together make up 35% of the total weight of the catch. He immediatelly sells these fish. At this point the 3 lightest fish make up 5/13 of the weight of the remaining fish.

    How many fish did the fisherman catch?
    Ten, not counting those he threw back.

  3. #33
    Join Date
    Apr 2007
    Posts
    762
    Quote Originally Posted by Homo Bibiens View Post
    8x6
    12x5

    l assume you do not consider 6x8 and 5x12 to be distinct from the above.

    Do you also allow 0x0?
    In general, where a=4(b-2)/(b-4) is a natural number for natural number b, the solution axb (or bxa) works.

  4. #34
    Join Date
    Sep 2007
    Location
    Lost among the 160,000+ members
    Posts
    8,002
    Quote Originally Posted by mr obvious View Post
    In general, where a=4(b-2)/(b-4) is a natural number for natural number b, the solution axb (or bxa) works.
    I got them all, though, right?

    a=4(b-2)/(b-4)

    so

    a=4+8/(b-4)

    So 8 has to be divisible by (b-4). So that means b=0 or b=2 (which are "paired" solutions, and leave a floor with an area of zero), or b=5, b=6, b=8, or b=12.

  5. #35
    Join Date
    Sep 2007
    Posts
    226
    Quote Originally Posted by Homo Bibiens View Post
    8x6
    12x5

    l assume you do not consider 6x8 and 5x12 to be distinct from the above.

    Do you also allow 0x0?
    8X6 and 12X5 does it for me. Well done!

  6. #36
    Join Date
    Sep 2007
    Posts
    226
    Quote Originally Posted by mr obvious View Post
    Ten, not counting those he threw back.
    Congratulations.

    Ten it is.

    Next question:

    Given that n is a whole number and that n^2 has 7 as its last but one digit, what is the last digit in n^2?

  7. #37
    Join Date
    Sep 2007
    Location
    Lost among the 160,000+ members
    Posts
    8,002
    6, and there are four different combinations of the last two digits in n that get it.

  8. #38
    Join Date
    Sep 2007
    Posts
    226
    Quote Originally Posted by Homo Bibiens View Post
    6, and there are four different combinations of the last two digits in n that get it.
    Correct!

    You have the honor of asking the next question.

  9. #39
    Join Date
    Sep 2007
    Location
    Lost among the 160,000+ members
    Posts
    8,002
    Quote Originally Posted by DanishDynamite View Post
    Correct!

    You have the honor of asking the next question.
    Uh oh, now I must think of a clever question. I displayed remarkable lack of foresight in choosing to answer the last question.

    Is it dishonorable or insulting in BAUT culture to give away my turn to anyone who likes to take it? I am still working on the airplane circumnavigation problem.

  10. #40
    Join Date
    Sep 2003
    Posts
    3,865
    Quote Originally Posted by Homo Bibiens View Post
    Is it dishonorable or insulting in BAUT culture to give away my turn to anyone who likes to take it?
    Not at all. I'll go next.

    My attempt, I remember, to remember ? finished in disaster, mate!

    What does ? represent?

  11. #41
    Join Date
    Sep 2007
    Posts
    363
    I don't understand the question?

  12. #42
    Join Date
    Dec 2005
    Posts
    1,905
    I understand the question!

    no wait...I don't.

  13. #43
    Join Date
    Apr 2006
    Posts
    4,181
    "I" is the only English word that could possibly fit ("a" doesn't quite work, neither does the contraction "o" ).

    Todd

  14. #44
    Join Date
    Apr 2007
    Posts
    762
    Does the ? need to be only a single letter?

  15. #45
    Join Date
    Sep 2003
    Posts
    3,865
    Quote Originally Posted by mr obvious View Post
    Does the ? need to be only a single letter?
    That's for me to know and for you to find out! It's a puzzle - but a mathematical one.

  16. #46
    Join Date
    Apr 2007
    Posts
    762
    Not sure what you mean. The number of letters in each word correspond to the digits of e, insofar as your sentence is a finite representation of the first few digits of e.

    So, the ? should be a single letter long. So, I guess tdvance was correct as far as the number of letters is concerned. But ?=e is probably better, except that that leads to a paradox since you clearly remember the mnemonic for e, you can determine quite a few digits and that would hardly constitute a disaster. So what's the rub?

  17. #47
    Join Date
    Sep 2003
    Posts
    3,865
    Quote Originally Posted by mr obvious View Post
    Not sure what you mean. The number of letters in each word correspond to the digits of e, insofar as your sentence is a finite representation of the first few digits of e.
    e is the correct answer.

    ... But ?=e is probably better, except that that leads to a paradox since you clearly remember the mnemonic for e, you can determine quite a few digits and that would hardly constitute a disaster. So what's the rub?
    Unfortunately success has only 7 letters! I was only a kid when I devised this mnemonic, so the illogic escaped me at the time. But it has served me well over the years.

    Over to you, mrobvious.

  18. #48
    Join Date
    Apr 2007
    Posts
    762
    Take a chessboard (8 by 8 grid) and remove one set of opposite corners (A1 and H8, for example, for those that know chess notation). There are 62 squares left. Take a single knight and place it anywhere on the board.
    (For those that don't know how knights move, see Diagram 2 here:
    http://en.wikipedia.org/wiki/Knight_(chess) )

    Your job is to describe a sequence of moves such that the knight visits every square exactly once (the square you start on counts as visited immediately), or explain why this is not possible.

  19. #49
    Join Date
    Sep 2003
    Posts
    3,865
    It should be impossible, since you removed two black squares, leaving 30 black and 32 white squares. When a knight moves, the squares it lands on are alternately black and white. So after visiting 30 black squares and 31 white squares, the knight will run out of black squares.

  20. #50
    Join Date
    Dec 2005
    Posts
    1,905
    I assume that is the correct answer.

    Next?

  21. #51
    Join Date
    Sep 2007
    Posts
    226
    There are exactly 2 proofs regarding number theory that I remember from my high school days. I hesitate to ask for either one as I'm sure they will be easily available using Google. Still, they are the only mathematical questions I can think of at this moment, so here's one of them:

    Prove that sqrt(2) is an irrational number.

  22. #52
    Join Date
    Apr 2007
    Posts
    762
    Let a/b = the rational expression of sqrt(2). Further, assume that a and b have gcd=1, or else reduce them so that their gcd =1.

    a=sqrt(2) * b
    a*a = 2*b*b (1)
    Thus, a must be even (if a is odd, then the '2' on the right can't divide a).
    let a = 2*n
    Substituting in equation 1,
    (2n)*(2n)=2*b*b
    4*n*n=2*b*b
    Dividing both sides by 2,
    2*n*n=b*b
    By the same reasoning as previously applied, b must be even. However, now both a and b are even, contradicting the initial condition that a and b must have gcd=1. Therefore, such a and b do not exist, therefore by definition sqrt(2) must be irrational.

  23. #53
    Join Date
    Sep 2007
    Posts
    226
    Quote Originally Posted by mr obvious View Post
    Let a/b = the rational expression of sqrt(2). Further, assume that a and b have gcd=1, or else reduce them so that their gcd =1.

    a=sqrt(2) * b
    a*a = 2*b*b (1)
    Thus, a must be even (if a is odd, then the '2' on the right can't divide a).
    let a = 2*n
    Substituting in equation 1,
    (2n)*(2n)=2*b*b
    4*n*n=2*b*b
    Dividing both sides by 2,
    2*n*n=b*b
    By the same reasoning as previously applied, b must be even. However, now both a and b are even, contradicting the initial condition that a and b must have gcd=1. Therefore, such a and b do not exist, therefore by definition sqrt(2) must be irrational.
    Congrats!

    Next question, please?

  24. #54
    Join Date
    Apr 2007
    Posts
    762
    For whatever reason, a king has decreed that he wishes to increase the female population in his kingdom. Without advanced medicine, he orders all families to stop having children once they have a son. His reasoning is that even though the chances of male vs. female in a given birth is 50-50, after a while, there will be families with one girl, two girls, three girls, etc. whereas there will be no families with more than one boy (and of course, there may be families with no boys - the people are not forced to have children until they have a boy; they simply can't have more once they do have a boy).

    Assuming this rule is extended to infinite time, that families have many children (e.g., no self-imposed abstinence), that natural resources are not a concern (this is a math problem, after all), what is the asymptotic ratio of females to males?
    Last edited by mr obvious; 2007-Sep-25 at 12:45 AM. Reason: typo, sorry

  25. #55
    Join Date
    Sep 2007
    Posts
    226
    Quote Originally Posted by mr obvious View Post
    For whatever reason, a king has decreed that he wishes to increase the female population in his kingdom. Without advanced medicine, he orders all families to stop having children once they have a son. His reasoning is that even though the chances of male vs. female in a given birth is 50-50, after a while, there will be families with one girl, two girls, three girls, etc. whereas there will be no families with more than one boy (and of course, there may be families with no boys - the people are not forced to have children until they have a boy; they simply can't have more once they do have a boy).

    Assuming this rule is extended to infinite time, that families have many children (e.g., no self-imposed abstinence), that natural resources are not a concern (this is a math problem, after all), what is the asymptotic ratio of females to males?
    Can a man have more than 1 wife (i.e. do surplus girls simply die as old spinsters or are they also impregnated?)

  26. #56
    Join Date
    Sep 2007
    Posts
    226
    Nevermind. It seems to me that the asymptotic ratio would still be 1.

  27. #57
    Join Date
    Apr 2007
    Posts
    762
    Correct, but an explanation would be nice

  28. #58
    Join Date
    Sep 2007
    Location
    Lost among the 160,000+ members
    Posts
    8,002
    If the probability of a boy is 1/2, then 1/2 of all families will have no girls, 1/4 of all families will have one girl, 1/8 of all families will have two girls, and so on. The expected number of girls is:

    0/(2^1)+1/(2^2)+2/(2^3)+3/(2^4)+...

    This series converges to one, so each family has children until it has one boy, but it also has one girl on average.

    It is simpler than that, though, if the probability of a boy is 1/2, then on average 1/2 of all children are boys, regardless of what rule families apply in when to stop.

  29. #59
    Join Date
    Apr 2007
    Posts
    762
    Fair enough. One of you two gets to ask a question. Please go ahead.

  30. #60
    Join Date
    Sep 2007
    Location
    Lost among the 160,000+ members
    Posts
    8,002
    DanishDynamite gave a correct answer, so probably he did everything correctly even though he did not explain. So I think it is his turn

Similar Threads

  1. Is the mainstream in physics being challenged?
    By Copernicus in forum Science and Technology
    Replies: 26
    Last Post: 2011-Sep-01, 06:30 AM
  2. why math?
    By tusenfem in forum Against the Mainstream
    Replies: 107
    Last Post: 2006-Apr-06, 07:19 AM
  3. Bad Math?
    By Sticks in forum Off-Topic Babbling
    Replies: 108
    Last Post: 2005-May-13, 02:14 AM
  4. Math Says NO!
    By Majin Vegeta in forum Against the Mainstream
    Replies: 113
    Last Post: 2005-Feb-22, 06:02 AM
  5. More math help
    By VTBoy in forum Off-Topic Babbling
    Replies: 24
    Last Post: 2004-Oct-29, 05:28 PM

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •