CAT 2020 Slot 1 QA Question 11: Among 100 students, $x_1$ have birthdays in January, $x_2$ have birthdays in February, and so on. If $x_0 = \max(x_1, x_2, \ldots, x_{12})$, then the smallest possible value of $x_0$ is

Question

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

Accepted Answer

To make the maximum value as small as possible, the 100 birthdays should be distributed as evenly as possible among the 12 months.

$\dfrac{100}{12} = 8$ remainder 4

So, 4 months can have 9 birthdays each and the remaining 8 months can have 8 birthdays each.

The maximum count is then 9. It cannot

Among 100 students, x1x_1x1​ have birthdays in January, x2x_2x2​ have birthdays in February, and so on. If x0=max⁡(x1,x2,…,x12)x_0 = \max(x_1, x_2, \ldots, x_{12})x0​=max(x1​,x2​,…,x12​), then the smallest possible value of x0x_0x0​ is

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