Among 100 students, \(x_1\) have birthdays in January, \(x_2\) have birthdays in February, and so on. If \(x_0\) = \(max(x_1,x_2,....,x_{12})\), then the smallest possible value of \(x_0\) is
Given 100 students, each born in one of 12 months. Let \( x_i \) be the count of students born in month \( i \), for \( i = 1, \ldots, 12 \).
Let \( x_0 = \max(x_1, x_2, \ldots, x_{12}) \). The objective is to find the minimum possible value of \( x_0 \).
Step 1: Aim for an even distribution.
Ideally, 100 students distributed over 12 months would mean \( \frac{100}{12} \approx 8.33 \) students per month. Since student counts are integers, some months will have 8 students and others will have 9.
Step 2: Determine the number of months with 8 or 9 students.
Let \( n \) be the number of months with 8 students. Then \( 12 - n \) months will have 9 students.
The total number of students is expressed as: \[ 8n + 9(12 - n) = 100 \]
Solving the equation: \[ 8n + 108 - 9n = 100 \\ -n = -8 \\ n = 8 \]
Step 3: Confirm the distribution.
- 8 months with 9 students: \( 8 \times 9 = 72 \)
- 4 months with 8 students: \( 4 \times 8 = 32 \)
Total students: \( 72 + 32 = 100 \) ✅
This distribution results in a maximum of 9 students in any given month. This is the smallest possible maximum.
\[ \boxed{9} \]