Among 100 students, x_1 have birthdays in January, x_2 have birthday in February, and so on. If x_0= (x_1, x_2, x_12), then the smallest possible value of x_0 is Option 1: 12 Option 2: 9 Option 3: 8 Option 4: 10

Answer: 9. Solution: We have 100 students whose birthdays are distributed across 12 months (January through December). We need to find the smallest possible value of the maximum number of students born in any single month. We think of it this way: If we want to minimize the "busiest" birthday month, we should try to spread the students as evenly as possible across all months. When we want to minimize the maximum value in a set, the best strategy is to make all values as close to each other as possible. Let's see what happens when we try to distribute 100 students equally across 12 months: 100 12 = 8.333... Since we can't have fractional students, we can't distribute them perfectly equally. We start with the base number: Each month gets at least: 100 12 = 8 students Total students accounted for: 8 x 12 = 96 students We distribute the remaining students: Remaining students: 100 - 96 = 4 students These 4 extra students must be distributed among some months We determine the final distribution: 8 months will have exactly 8 students each 4 months will have exactly 9 students each (8 + 1 extra) We have: 8 x 8 + 4 x 9 = 64 + 36 = 100 Could we do better than 9? Let's check if we could make the maximum 8 or less: If every month had at most 8 students: 8 x 12 = 96 students But we need to place 100 students, so this is impossible! Could we do worse? If we distributed unevenly (like putting many students in one month), the maximum would be higher than 9. The smallest possible value of x_0 (the maximum number of students born in any month) is 9. This occurs when we distribute students as evenly as possible: 8 months with 8 students each, and 4 months with 9 students each.

Algebra

Minima & Maxima

Easy

Among $100$ students, $x_{1}$ have birthdays in January, $x_{2}$ have birthday in February, and so on.

If $x_{0}=\max \left(x_{1}, x_{2}, \ldots , x_{12}\right)$ , then the smallest possible value of $x_{0}$ is

$12$

$9$

$8$

$10$

✅ Correct Option: 2

We have 100 students whose birthdays are distributed across 12 months (January through December). We need to find the smallest possible value of the maximum number of students born in any single month.

We think of it this way: If we want to minimize the "busiest" birthday month, we should try to spread the students as evenly as possible across all months.

When we want to minimize the maximum value in a set, the best strategy is to make all values as close to each other as possible.

Let's see what happens when we try to distribute 100 students equally across 12 months:

$100 \div 12 = 8.333...$

Since we can't have fractional students, we can't distribute them perfectly equally.

We start with the base number:

Each month gets at least: $\lfloor 100 \div 12 \rfloor = 8$ students

Total students accounted for: $8 \times 12 = 96$ students

We distribute the remaining students:

Remaining students: $100 - 96 = 4$ students

These 4 extra students must be distributed among some months

We determine the final distribution:

8 months will have exactly 8 students each

4 months will have exactly 9 students each (8 + 1 extra)

We have: $8 \times 8 + 4 \times 9 = 64 + 36 = 100$

Could we do better than 9? Let's check if we could make the maximum 8 or less:

If every month had at most 8 students: $8 \times 12 = 96$ students

But we need to place 100 students, so this is impossible!

Could we do worse? If we distributed unevenly (like putting many students in one month), the maximum would be higher than 9.

The smallest possible value of $x_0$ (the maximum number of students born in any month) is 9.

This occurs when we distribute students as evenly as possible: 8 months with 8 students each, and 4 months with 9 students each.

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CAT Past Year Questions (Topic-Wise)

Among $100$ students, $x_{1}$ have birthdays in January, $x_{2}$ have birthday in February, and so on.

If $x_{0}=\max \left(x_{1}, x_{2}, \ldots , x_{12}\right)$ , then the smallest possible value of $x_{0}$ is