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Among $100$ students, $x_{1}$ have birthdays in January, $x_{2}$ have birthdays in February, and so on. If $x_{0}= \text{max}\left ( x_{1},x_{2},\dots,x_{12} \right ),$ then the smallest possible value of $x_{0}$ is

1. $9$
2. $10$
3. $8$
4. $12$

Given that, $x_{0} = \text{max} (x_{1}, x_{2}, \dots, x_{12})$

That means,

• $x_{0} \geq x_{1}$
• $x_{0} \geq x_{2}$
• $x_{0} \geq x_{3}$
• $x_{0} \geq x_{4}$
• $x_{0} \geq x_{5}$
• $x_{0} \geq x_{6}$
• $x_{0} \geq x_{7}$
• $x_{0} \geq x_{8}$
• $x_{0} \geq x_{9}$
• $x_{0} \geq x_{10}$
• $x_{0} \geq x_{11}$
• $x_{0} \geq x_{12}$

Now, $x_{1} + x_{2} + \dots + x_{12} = 100$

$\Rightarrow x_{0} + x_{0} + \dots + x_{0} \geq 100$

$\Rightarrow 12 x_{0} \geq 100$

$\Rightarrow x_{0} \geq \frac{100}{12}$

$\Rightarrow x_{0} = \left \lceil \frac{100}{12} \right \rceil$

$\Rightarrow x_{0} = \left \lceil 8.33 \right \rceil$

$\Rightarrow \boxed{x_{0} = 9}$

Correct Answer$: 9$

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