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Brain Teasers and Math Problems

  1. Apr 11, 2006 #1

    siddharth

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  2. jcsd
  3. Apr 11, 2006 #2
    Is it just me, or is this incorrect? I wrote a script that started Jan 13th, 1970, and went through until Dec 13th, 2037, and came up with the following counts for 13ths of the month:

    Mon - 116
    Tue - 118
    Fri - 117
    Wed - 115
    Sun - 117
    Thu - 117
    Sat - 116

    Am I missing something? Do I need a far more significant dataset before the trend becomes visible?

    DaveE
     
  4. Apr 11, 2006 #3

    NateTG

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    You'll want to run through at least 400 years to account for leap year cycles.
    Since 7 divides the number of days in 400 consecutive years (365*400+97), that's a complete cycle.
     
  5. Apr 11, 2006 #4
    Yep, THAT did it, thanks! Had to modify the program to include leap year logic rather than use existing functions. Ran it on 146097 days (the complete cycle) and came up with:

    Mon - 685
    Tue - 685
    Fri - 688
    Wed - 687
    Sun - 687
    Sat - 684
    Thu - 684

    DaveE
     
  6. Apr 11, 2006 #5
    I got a few of the quickies.

    4. [itex]-1 = e^{i\pi}[/itex]
    5. [itex]\left(\frac{499}{500}\right)^{500}[/itex]
    9. o{ne}, t{wo}, t{hree}, f{our}, f{ive}, s{ix}, s{even}, e{ight} and so the answer is n, t, e, t, t, f, etc.
    16. Footnote 4 is the same as footnote 3, but in English.

    P.S.
    I don't know why the first two answers aren't hidden.
     
    Last edited: Apr 11, 2006
  7. Apr 11, 2006 #6
    I'm not sure if [tex]-i\pi[/tex] counts as irrational, so I decided to give an alternate demonstration:

    We know that [tex] 2 = e^x[/tex] for some x. If x were rational (say x= p/q clearly neq 0), then we have [tex]2^q=e^p[/tex] but no positive integer power of e can be an integer, so this is a contradiction.
    (If we did have [tex] n = e^p[/tex] for positive integers n,p, then e would be a root of [tex] x^p -n \in \mathbb{Q}[x][/tex])
     
  8. Apr 12, 2006 #7

    NateTG

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    Resorting to 'e is transcendental' without proof is IMHO unsatisfying.

    I'd prefer something like:
    Consider the real solution to:
    [tex]x^x=2[/tex]

    Clearly, [itex]x[/itex] is not an integer. Now, if [itex]x[/itex] is rational, then we have:
    [tex]x=\frac{a}{b}[/tex]
    expressed as a fraction in lowest terms, and with [itex]b\neq1[/itex].
    But then we can do the following:
    [tex](x^x)^b=2^b[/tex]
    [tex]\left(\frac{a}{b}\right)^a=2^b[/tex]

    Since [itex]a[/itex] and [itex]b[/itex] are integers, we have the LHS is a non-integer, and the RHS is, which is a contradiction - so the assumption that [itex]x[/itex] is rational is clearly false.
     
  9. Apr 12, 2006 #8

    VietDao29

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    For Quicky #10:
    Here are sets of numbers enclosed by square brackets. Fill the next set of numbers. [1],[1,1],[2,1],[1,2,1,1],[1,1,1,2,2,1],[3,1,2,2,1,1], [.............]
    Start with [1], we count 1 number 1, so we write [1, 1], then we continue counting 2 numbers 1, so we write [2, 1], now that's 1 number 2, and 1 number 1, so we write [1, 2, 1, 1],...
    So a few next sets are:
    [1, 1, 1, 2, 2, 1]
    [3, 1, 2, 2, 1, 1]
    [1, 3, 1, 1, 2, 2, 2, 1]
    [1, 1, 1, 3, 2, 1, 3, 2, 1, 1]
    ...
    For # 15,
    1023 players (yes, not 1024) participate in a tournament in which each game produces a decisive winner. Players are eliminated by knock-out with byes being given when odd number of players occur at any given round. How many matches need to be played to find a winner ?
    I think it's 1022, since 1022 playes must be knocked-out before the winner is found.
     
  10. Apr 12, 2006 #9
    The answer to quickie #11 is yes
    The answer to quickie #12 is use base 15
     
  11. Apr 12, 2006 #10

    siddharth

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    jimmy,
    how is the fifth one [itex]\left(\frac{499}{500}\right)^{500}[/itex]?
    Isn't that very high?
     
  12. Apr 12, 2006 #11
    Answer to quickie no 5: Probability of 0 typos = 499/999
    Answer to quickie no 13:2.3 , 2ln3
    Didn't find time 4 the rest.
    Nice collection anyways.

    --Arun
     
    Last edited: Apr 12, 2006
  13. Apr 12, 2006 #12
    Answer to quickie #3 is 1001! + 2, 1001! + 3, ... 1001! + 1001
     
  14. Apr 12, 2006 #13
    It's about 0.3675
    Suppose there were only 1 error in the book. Then what would the answer be? How about 2 errors? Since the errors are randomly distributed, the probablility of both errors not being on page 29 is the product of the probablilities that either one is not on page 29. How about 500 errors?
     
  15. Apr 12, 2006 #14

    NateTG

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    You're making some assumptions about the distribution of errors that aren't necessarily true.
     
  16. Apr 12, 2006 #15
    I don't believe so.
     
  17. Apr 12, 2006 #16

    siddharth

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    Answer to quickie #8 is staring right out of the keyboard
     
  18. Apr 12, 2006 #17
    What jimmy is trying to ask is how the probability will be affected if the no of errors were different from the number of pages.
    However I feel his answer is wrong.
    Let me explain how I arrived at my answer( which can be extended to general cases)
    The general formula for distributing n things among r different people is given by
    [tex]^{n+r-1}C_{r-1}[/tex]

    For this case,the no of favourable outcomes (no error in page 29) from the above formula will be [itex]^{998}C_{498}[/itex] (distributing 500 errors in 499 pages leaving page 29 vacant).
    No of possible outcomes will be [itex]^{999}C_{499}[/itex](distributing 500 errors in 500 pages)
    So the probability will be their ratio which turns out to be 499/999
    (slightly less than half).
    The same procedure may be applied for other variations od the problem.

    - Arun
     
  19. Apr 12, 2006 #18

    NateTG

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    I agree that if one assumes that the typos are independantly assigned with an even random distribution, then the probability that page 29 has no typos is going to be
    [tex]\left(\frac{499}{500}\right)^{500}[/tex]

    However, consider, for example, a book that has 1 character per page, 500 pages, and 500 typos. The probability that a typo occurs on page 29 is then 1.
     
  20. Apr 12, 2006 #19
    For this you receive an AAA+.

    Edit: Sorry, make that an A+.
     
  21. Apr 12, 2006 #20
    Try this out with 2 errors distributed between 2 pages. I get 1/4 as the probability that page 1 is free of errors in two different ways:

    method 1)
    error 1 on page 1, error 2 on page 1: page 1 is not error free
    error 1 on page 1, error 2 on page 2: page 1 is not error free
    error 1 on page 2, error 2 on page 1: page 1 is not error free
    error 1 on page 2, error 2 on page 2: page 1 is error free
    probability that page 1 is error free: 1/4

    method 2)
    [tex]\left(\frac{n-1}{n}\right)^n[/tex] where n = 2
    probability that page 1 is error free: 1/4

    Using your method, I get)
    [tex]\frac{^{n+r-1}C_{r-1}}{^{n+r}C_{r}}[/tex]where n = 2 and r = 1
    probability that page 1 is error free: 1/3

    I think your method is wrong. I think it counts both instances of 1 error per page as if they were a single instance.
     
    Last edited: Apr 12, 2006
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