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

We are given that $x_1+x_2+\dots +x_{12}=100$ for the 12 months of the year.

$x_0$ represents the maximum value among all these monthly values. We want to find the smallest possible value that this maximum value $x_0$ can take.

To minimize the maximum value of a set of integers summing to a fixed constant, we must distribute the values as evenly as possible.
Dividing 100 by 12:
$100=12\times 8+4$

This means we can have 8 months with 8 birthdays and 4 months with 9 birthdays (since $8\times 8+9\times 4=64+36=100$).
In this case, the maximum value $x_0=max(8,8,\dots ,9,9)=9$.

If we try to make $x_0$ smaller, say 8, then the maximum number of birthdays in any month would be 8. In that case, the total sum of birthdays across 12 months could be at most:
$12\times 8=96$
But the total sum must be 100, which is a contradiction. Therefore, the minimum possible value of the maximum is 9.