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.
Best Answer
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?