Find the smallest positive integer $n$ satisfying the following two properties:

1. $n$ has exactly 144 distinct positive divisors,
2. among the positive divisors of $n$, there are ten consecutive integers.

Go to the cell corresponding to the integer part of $\sqrt{n}$ to find a puzzle.
Among 10 consecutive integers, there is necessarily a multiple of 8, one of 9, one of 5, and one of 7. Therefore, $n$ is at least divisible by $2^{3} \cdot 3^{2} \cdot 5 \cdot 7$. Let's write the prime factorization of $n$:

$$
n=2^{\alpha_{2}} 3^{\alpha_{3}} 5^{\alpha_{5}} 7^{\alpha_{7}} 11^{\alpha_{11}} \ldots
$$

The number of divisors of $n$ is then

$$
\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{5}+\right)\left(\alpha_{7}+1\right)\left(\alpha_{11}+1\right) \cdots \geq 48 \cdot\left(\alpha_{11}+1\right) \ldots
$$

This quantity equals 144, so $\left(\alpha_{11}+1\right) \cdots \leq 3$, meaning there is at most one non-zero $\alpha_{p}$ for the $p$ greater than 5, and for the sake of minimality, it will be $\alpha_{11}$. After some study, we arrive at the following two possibilities: $n=2^{5} 3^{3} 5^{2} 7=151200$ and $n=2^{3} 3^{2} 5^{2} 7 \cdot 11=110880$. The second one is the best.