Smallest Possible Value

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.

• Pigeonhole principle? Commented Dec 24, 2020 at 16:43
• Hint: $\max(\cdots) \ge {\rm avg}(\cdots)$ and $x_0$ is an integer. Commented Dec 24, 2020 at 16:44
• 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) Commented Dec 24, 2020 at 16:48

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?
• Just a silly correction: $100/12 \sim 8.33$ Commented Dec 24, 2020 at 16:50