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I was given this question. Maths is a weakness of mine and I'm really struggling to understand what this question is asking, and how to solve it. My notes indicate it's a pigeon hole principle but I'm unsure.

"You prepare a schedule of your physical exercises for the next fortnight (14 days). You don’t do physical exercise more than once a day. If you have 10 PE sessions planned, explain using the counting principles covered in class how this means you will do PE on consecutive days at least once in the next fortnight."

Thanks in advance :)

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Let $\ n\ $ be the number of PE lessons scheduled for days immediately followed by a day free of any PE lesson. These lessons plus the following lesson-free day must occupy a total of $\ 2n\ $ days, at most one of which (if it is the day following the last lesson) can lie outside the allowed $14$-day period. This leaves at most $\ 15-2n\ $ days available to be occupied by the remaining $\ 10-n\ $ lessons. Thus, $\ 10-n\le15-2n\ $, giving $\ n\le5\ $. Thus, there must be at least $\ 5\ $ lessons on days that are immediately followed by a day with another lesson.

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  • $\begingroup$ And how exactly do I apply these values to this formula "For natural numbers k and m, if n = km + 1 objects are distributed among m sets, then at least one of the sets will contain k + 1 or ceiling(n/m) objects." $\endgroup$
    – Newb
    Commented Oct 30, 2020 at 2:27
  • $\begingroup$ The form of the principle which you need is "If $\ n\ $ objects are distributed among $\ m\ $ sets, then at least one set must contain at least $\ \left\lceil\frac{n}{m}\right\rceil\ $ objects." If you pair up successive days, as PiKindOfGuy suggests in his answer, then you have $\ n=10\ $ objects which have to be distributed among the $\ m=7\ $ "sets" (i.e. pairs of days). So at least one such pair must contain at least $\ \left\lceil\frac{10}{7}\right\rceil=2\ $ objects. $\endgroup$ Commented Oct 30, 2020 at 9:16
  • $\begingroup$ Note, however, that $\ n\ $ is not of the form $\ mk+1\ $, but $\ mk+3\ $, where $\ k=1\ $. $\endgroup$ Commented Oct 30, 2020 at 12:50
  • $\begingroup$ That's very helpful! Thanks heaps! How come my formula says + 1 if it's '3' in this example? is the '1' just a placeholder for any integer value? $\endgroup$
    – Newb
    Commented Oct 31, 2020 at 0:22
  • $\begingroup$ I don't know. Where did you get it from? As you've stated it, it contains a potentially misleading ambiguity, and is of limited applicability. The ambiguity is in the clause "will contain $\ k+1\ $ ... ," which I would normally take to mean "will contain exactly $\ k+1\ $ ... ". But that would make the statement incorrect. For the statement to be correct this clause has to be interpreted as meaning "will contain at least $\ k+1\ $ ... ". Also, the statement is only applicable when $\ n\equiv 1\pmod{m}\ $, as $\ n\ $ could not otherwise have the form $\ km+1\ $. $\endgroup$ Commented Oct 31, 2020 at 5:15

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