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    A football team is playing in the off-season, which lasts 77 days. The coach wants the team to play at least one practice game each day, but no more than 132 games over the course of the off-season. Show that there is a period of consecutive days in which the team plays exactly 21 games.
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    (Original post by theone)
    Rep to the first person to solve this
    Ooooo I am motivated..but I don't have the patience for this.
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    (Original post by theone)
    Rep to the first person to solve this

    A football team is playing in the off-season, which lasts 77 days. The coach wants the team to play at least one practice game each day, but no more than 132 games over the course of the off-season. Show that there is a period of consecutive days in which the team plays exactly 21 games.
    you what?
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    (Original post by elpaw)
    you what?
    :rolleyes:
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    (Original post by theone)
    :rolleyes:
    no, im just confused by the question.
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    (Original post by theone)
    :rolleyes:
    Is it possible? It seems very hard, it could be a piece of coursework..
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    (Original post by elpaw)
    no, im just confused by the question.
    There's 77 days, during eahc of which one game at least must be played. But there can not be more than 132 games played over the whole 77 days. Show that we can find a period of consecutive days in which exactly 21 games are played.
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    (Original post by theone)
    There's 77 days, during eahc of which one game at least must be played. But there can not be more than 132 games played over the whole 77 days. Show that we can find a period of consecutive days in which exactly 21 games are played.
    I still don't get exactly what you mean. But I think I got some sort of idea. Peogeon hole principle?

    132 / 77 = 1 remainder 55? yeah?
    and that means you can have 55 days which have two games played (hepothetical situation). and you can then take 10 days from the 55 then add on to it one day which has only one game played. 21 games in 11 days.

    situation 2, suppose we have 1 game played on the 1st day and 2 games played on the 2nd day and so on. But we don't want to see 21 games in consecutive days.

    1, 2, 3, 4, 5, 5 --- 20 games in 6 days, 77 / 6 = 12 remaider whatever. but 20 * 12 = 240> 132 therefore impossible.

    I can't think anything right now but I think this is how you solve it. Hope this helps.

    (Original post by gemgems89)
    Is it possible? It seems very hard, it could be a piece of coursework..
    Ans: Very possible and may be "travial" in some people's eyes. This can never be a piece of coursework. And I think he's preparing for Wednesday's big Exam.
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    (Original post by theone)
    Rep to the first person to solve this
    You must have solved it, haven't you. And now you are just asking other people who can't solve it to feel better? Are you? I hope not.

    How many peogeons are there in the hole?
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    If we say the number of games played up to and including day n, we can say that:

    1<=P1<P2<P3<........<P77

    and also P77=132.

    Add 21 on to each of these terms to get:

    22<=P1+21<P2+21<P3+21<...<P77+21 =153

    I now have a list of 154 numbers which have integer values between 1 and 153 inclusive. Therefore, there must be at least a pair of numbers which are the same.

    However, Pn cannot equal Pk where k>n because P1<P2<P3 and so on. Also, Pn+21 cannot equal Pk+21 where k>n because P1+21<P2+21 and so on.

    Therefore, Pn=Pk+21. Thus, there exists a day where 21 more matches have been played up to and including it than another day. Therefore, there must be a series of consecutive days when 21 games are played.

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    (Original post by meepmeep)
    If we say the number of games played up to and including day n, we can say that:

    1<=P1<P2<P3<........<P77

    and also P77=132.

    Add 21 on to each of these terms to get:

    22<=P1+21<P2+21<P3+21<...<P77+21 =153

    I now have a list of 154 numbers which have integer values between 1 and 153 inclusive. Therefore, there must be at least a pair of numbers which are the same.

    However, Pn cannot equal Pk where k>n because P1<P2<P3 and so on. Also, Pn+21 cannot equal Pk+21 where k>n because P1+21<P2+21 and so on.

    Therefore, Pn=Pk+21. Thus, there exists a day where 21 more matches have been played up to and including it than another day. Therefore, there must be a series of consecutive days when 21 games are played.

    Good enough for the rep! Nice trick.

    And I hadn't actually done all of it.
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    (Original post by theone)
    Good enough for the rep! Nice trick.

    And I hadn't actually done all of it.
    Thanks. Some nice practice for the BMO on Wednesday (and the far scarier prospect of the Cambridge interview one week after!)
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    (Original post by meepmeep)
    However, Pn cannot equal Pk
    whats Pk?
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    (Original post by edders)
    whats Pk?
    k is just another day later than n, so Pk is the number of matches played up to and including day k.
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    To theone:
    Where did you find this question? I'm asking because yesterday we had a maths test at school and they gave us this question. I think nobody solved it correctly. Anyway,I would appreciate it, if you could tell me where you found it.
    Thanks.
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    (Original post by meepmeep)
    k is just another day later than n, so Pk is the number of matches played up to and including day k.
    um right :confused:
    i still dont get the reasoning lol. ah well looks like im going to fail my oxford interview
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    The proof basically says that there will be a day when 21 more games up to and including it are played than another day (so 21 more games have been played on consecutive days).

    Having read that, not sure it makes it any clearer, but nevermind.
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    (Original post by meepmeep)
    The proof basically says that there will be a day when 21 more games up to and including it are played than another day (so 21 more games have been played on consecutive days).

    Having read that, not sure it makes it any clearer, but nevermind.
    oh right i think i get you now... you could do much the same with any other number eg. 20 surely?
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    Yeah, but it doesn't work if you need to play 22 games on consecutive days, as you'd get 154 numbers which are integers which have a value between 1 and 154 inclusive, so a pair does not have to occur.
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    (Original post by meepmeep)
    Yeah, but it doesn't work if you need to play 22 games on consecutive days, as you'd get 154 numbers which are integers which have a value between 1 and 154 inclusive, so a pair does not have to occur.
    whats this talk of pairs? where does that come into it?

    ive given you rep for stimulating my brain btw
 
 
 
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