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Among $100$ students, $x_1$ have birthdays in January, $x_2$ have birthdays in February, and so on. If $x_0 = \max(x_1, x_2, \dots, x_{12})$, then the smallest possible value of $x_0$ is...

I came across this question and have been trying to get my head around it and the way to solve this. I am not able to think of the way this question should be approached. A little help or hint would be very helpful.

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  • $\begingroup$ Pigeonhole principle? $\endgroup$
    – Rivers McForge
    Commented Dec 24, 2020 at 16:43
  • $\begingroup$ Hint: $\max(\cdots) \ge {\rm avg}(\cdots)$ and $x_0$ is an integer. $\endgroup$
    – achille hui
    Commented Dec 24, 2020 at 16:44
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    $\begingroup$ I'm not very sure I understood correctly the question.. If $x_1$ is for example the number of people birth in Jenuary and so on and they do not have any distribution then the smallest possible value for $x_0$ = $\lceil 100/12 \rceil$ and this is the case where all the months have the same births (execpted for some that must end up with 1 birthday more than the others) $\endgroup$
    – Gabrielek
    Commented Dec 24, 2020 at 16:48

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Note that $x_0 \ge x_1$, $x_0 \ge x_2$, ... , $x_0 \ge x_{12}$. Thus

$$100=x_1 + x_2 + \cdots + x_{12} \le x_0 + x_0 + \cdots + x_0 = 12 x_0$$ This shows that $$100 \le 12 x_0$$ This gives us a lower bound: $x_0$ is at least $100/12 \approx 8.6$, i.e. $x_0$ is at least $9$.

Now you have to show that $x_0$ could be $9$. Can you conclude from here?

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    $\begingroup$ Just a silly correction: $100/12 \sim 8.33$ $\endgroup$
    – Gabrielek
    Commented Dec 24, 2020 at 16:50

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