Question

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

• A. 9
• B. 10
• C. 8
• D. 12

Answer 1

The Correct Option is A

We are given that there are 100 students, each born in one of the 12 months. Let: \[ x_1, x_2, \ldots, x_{12} \] represent the number of students born in each month.

Let \( x_0 = \max(x_1, x_2, \ldots, x_{12}) \), the maximum number of students in any one month. Our goal is to minimize \( x_0 \).

Step 1: Distribute 100 students as evenly as possible over 12 months.

If the distribution were perfectly even: \[ \frac{100}{12} \approx 8.33 \] But since student counts must be integers, some months will have 8 students, and some will have 9.

Step 2: Let \( n \) be the number of months with 8 students.
Then the remaining \( 12 - n \) months will have 9 students.

So the total number of students is: \[ 8n + 9(12 - n) = 100 \]

Expanding and simplifying: \[ 8n + 108 - 9n = 100 \\ -n + 108 = 100 \\ n = 8 \]

Step 3: Verify the distribution:
- 8 months have 9 students → \( 8 \times 9 = 72 \)
- 4 months have 8 students → \( 4 \times 8 = 32 \)
Total = \( 72 + 32 = 100 \) ✅

Therefore, the maximum number of students in any month is: \[ \boxed{9} \] which is the smallest possible maximum.