Math challenges II: Want to give it a try?

Discussion in 'General Education Archives' started by Carmen13, May 6, 2006.

  1. Carmen13

    Carmen13 Groupie

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    May 10, 2006

    Yes, you are right TeacherGroupie! Not the neper number, just number 3!
    You're welcome. Glad you found it fun!:angel:
     
  2. TeacherGroupie

    TeacherGroupie Moderator

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    Yes indeed, at least once I got done cussing...
     
  3. Carmen13

    Carmen13 Groupie

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    May 10, 2006

    Extra questions::rolleyes:

    1) How many persons stay in line, through each step, until the winners are selected?

    2) If the number of persons in line, initially, is bigger than 1458 and smaller than 6562 (exclusive), will we have the same winners?
     
  4. Upsadaisy

    Upsadaisy Moderator

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    Waaaahhhh .... I want a new problem.
     
  5. Carmen13

    Carmen13 Groupie

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    okay, okay! Wait a sec...:p
     
  6. Carmen13

    Carmen13 Groupie

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    May 10, 2006

    :eek: What a joke!!!

    I decided to tell a joke to: one person, today, and two persons, tomorrow.
    I will ask, each one of those persons, to tell the joke to: one person, the following day to the day they have heard the joke, and two persons, the day after the following day {that is, two days after they heard the joke}. I will also ask them to make this same request, to the 3 persons they will tell the joke to.
    By the end of one week, how many persons (at the most), will know the joke?

    PS: At least the titles are a creation of mine!:D

    (Portuguese Math Olympics 2005, 16st of March 2005, grades 8/9)
     
  7. Upsadaisy

    Upsadaisy Moderator

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    21?
     
  8. Carmen13

    Carmen13 Groupie

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    May 11, 2006

    No...
     
  9. mshutchinson

    mshutchinson Comrade

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    May 11, 2006

    I didn't get the formula..
    Can you explain it in English?
    Should we have nkown that 3 would be raised to the 6th power??
     
  10. Carmen13

    Carmen13 Groupie

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    May 11, 2006

    1. In each step, the difference between each number and the previous one (let us call it d) is a power of 3:

    step 1: 1,4,7,10,...
    d=3

    step 2: 1,10,19,...
    d=9=3^2

    ...

    step k: 1. 3^k+1, 2* 3^k+1, 3*3^k+1, ...
    d=3^k


    2. So we conclude that

    - the first winner is number 1;

    - the expression for the second winner is 3^k +1;

    - the expression for the third winner is (3^k +1)+3^k =2 * 3^k +1.

    3. We know that the last step is k=6 because

    2* 3^k +1<2005 and 3*3^k+1>2005

    therefore

    668<3^k<1002

    which implies k=6.

    _________________________

    Note: There are other ways of determining the number of steps...
     
  11. TeacherGroupie

    TeacherGroupie Moderator

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    May 11, 2006

    Okay, so before the first pass, the people are 1, 2, 3, etc., and the difference between one and the next is 1.

    After the first pass, the people left are

    1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, 97, 100, 103, 106, 109...

    The difference between 1 and 4 is 3, as is the difference between 4 and 7, 7 and 10, 10 and 13, etc.

    After the second pass, the people left are

    1, 10, 19, 28, 37, 46, 55, 64, 73, 82, 91, 100, 109...

    The difference this time is 9. Hm: that could be 3 + 6, somehow, or it could be 3 x 3.

    After the third pass, the people left are

    1, 28, 55, 82, 109...

    The difference this time is 27. Oho: 27 is 3 cubed - 3 to the THIRD power, and this is the THIRD pass; and on the SECOND pass the difference was 3 to the SECOND power; and on the FIRST pass, the difference was 3 to the FIRST power (otherwise known as plain old 3); and before the first pass the difference was 1, and it turns out that 3 raised to the 0 power is 1. A pattern!

    The question didn't actually ASK whether in this case the pattern required 3 to the SIXTH power... but this is where, in addition to listing, we start calculating 3^4, then 3^5, etc., and adding and checking till it's clear that the fourth term would be greater than 2005.
     
  12. TeacherGroupie

    TeacherGroupie Moderator

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    May 11, 2006

    Ah: Carmen and I came up with somewhat different approaches, but they do indeed lead to the same end. I will sit here and cheerfully bask in reflected glory, since I'm not half the mathematician Carmen is.
     
  13. Carmen13

    Carmen13 Groupie

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    May 11, 2006

    Well done TG!:angel:
     
  14. TeacherGroupie

    TeacherGroupie Moderator

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    May 11, 2006

    I should add that amazingly many number patterns involve exponents of 2 and 3. The prudent student on seeing 8, for example, recalls that 2^3 is at least a possibility.
     
  15. SportsFanTr

    SportsFanTr Companion

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    May 11, 2006

    39?
     
  16. Carmen13

    Carmen13 Groupie

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    May 11, 2006

    No... the solution is bigger than 100.
     
  17. tikibad

    tikibad Rookie

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    May 12, 2006

    The first person (number 1)
    The middle person (number 1003)
    The last person (number 2005)
     
  18. tikibad

    tikibad Rookie

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    May 12, 2006

    Oh my gosh... I was on the first page... so I posted for the first page... what an idiot.

    Okay.... lemme read the last one.
     
  19. Carmen13

    Carmen13 Groupie

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    May 12, 2006

    Tikibad, your solution is not correct...the solution and explanation has already been presented....Solve it again.
    You can also try the next problem!
     
  20. tikibad

    tikibad Rookie

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    May 12, 2006

    Is there a quicker way

    I had to draw out a diagram

    The answer is 165 (including yourself)

    Is there a quicker way, other than drawing a diagram.
     
  21. Carmen13

    Carmen13 Groupie

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    May 12, 2006

    Try organizing the data in a table and finding a formula to obtain the number of persons who are told the joke, per day...
    oh, and the answer is not 165... it is close though!
     
  22. Carmen13

    Carmen13 Groupie

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    I started with a diagram, but gave up soon enough!:p
     
  23. tikibad

    tikibad Rookie

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    sorry... miss counted... 164
     
  24. Carmen13

    Carmen13 Groupie

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    May 13, 2006

    :sorry: but no...the solution is 170.
    I would love if someone presented the resolution...:love:
     
  25. Carmen13

    Carmen13 Groupie

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    May 15, 2006

    Hint...


    Day | Total per day | Total

    1 | 1 | 1

    2 | 1+2 =3 | 4

    3 | 3*1 + 1*2 = 5 | 9
     
  26. Carmen13

    Carmen13 Groupie

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    May 17, 2006

    Day | Total per day | Total

    4 | 5*1+3*2 = 11| 20

    5 | 11*1+5*2 = 21| 41

    6 | 21*1+11*2 = 43| 84

    7 | 43*1+21*2 = 85| 169

    169 + 1 (me) = 170 persons
     
  27. Carmen13

    Carmen13 Groupie

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    May 18, 2006

    :eek: MONSTER number!

    Andreia wrote all the integer numbers from 1 to 2002, in increasing order.
    She obtained the MONSTER number

    123456789101112131415...19981999200020012002.​

    What is its central digit?
     
  28. Upsadaisy

    Upsadaisy Moderator

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    0 ?
     
  29. Carmen13

    Carmen13 Groupie

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    May 19, 2006

    No...
    Upsadaisy, what strategy did you use?
     
  30. Upsadaisy

    Upsadaisy Moderator

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    Why isn't it just 1001 in the middle?
     
  31. Carmen13

    Carmen13 Groupie

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    Upsadaisy, the MONSTER number is a number, not a sequence of numbers.
    Start by finding the number of digits of this MONSTER...:eek:
     
  32. Upsadaisy

    Upsadaisy Moderator

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    You monster!!!
     
  33. tikibad

    tikibad Rookie

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    May 25, 2006

    1 digit = 9 * 1 = 9
    2 digit = 90 * 2 = 180
    3 digit = 900 * 3 = 2700
    4 digit = 1003 * 4 = 4012

    4012 + 2700 + 180 + 9 = 6901 digits

    6900 divided by 2 = 3450

    digit # 3451 is the central digit.

    *** 9 (1 digit) + 180 (2 digit) + 2700 (3 digit) = 2889

    3450 - 2889 = 561 divided by 4 (4 digit numbers) = 140.25

    Anyway... the central number is .....................

    1
     
  34. Carmen13

    Carmen13 Groupie

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    Well done tikibad!:angel:

    561=140*4+1

    so 1000+140=1140
    _______________
    Note that, if we had obtained 562, for instance, then

    562=140*4+2, hence the central digit would have been 4.
    1000+140=1140.
     
  35. tikibad

    tikibad Rookie

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    May 26, 2006

    Actually... it's 999 + 140... because 1000 is a 4 digit number...
     
  36. Carmen13

    Carmen13 Groupie

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    Jun 1, 2006

    :eek: A cool bike

    What's the area of the painted parts?

    [​IMG]

    (Created by carmen13, using Geometer's sketchpad)
     
  37. TeacherGroupie

    TeacherGroupie Moderator

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    Stinker!
     
  38. tikibad

    tikibad Rookie

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    Jun 2, 2006

    10 meters 77 centimeters ... ABOUT!
     
  39. TeacherGroupie

    TeacherGroupie Moderator

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    Um, tikibad... the last time I checked, area wasn't in meters, it was in SQUARE meters.
     
  40. Carmen13

    Carmen13 Groupie

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    :confused: :thanks:
     

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