2. Find the smallest positive integer $n$ that satisfies: (1) $n$ has exactly 144 different positive divisors; (2) Among the positive divisors of $n$, there are 10 consecutive integers.
2. Solution: Since in 10 consecutive positive integers, there must be numbers divisible by $2^{3}, 3^{2}, 5, 7$.

Therefore, $n=2^{a_{1}} \cdot 3^{a_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4}} \cdot 11^{a_{5}} \cdots$ where, $a_{1} \geqslant 3, a_{2} \geqslant 2, a_{3} \geqslant 1, a_{1} \geqslant 1, \cdots$
Since the number of positive divisors of $n$ is $\left(a_{1}+1\right)\left(a_{2}+1\right) \cdots=144$,
and $\left(a_{1}+1\right)\left(a_{2}+1\right)\left(a_{3}+1\right)\left(a_{4}+1\right) \geqslant 4 \cdot 3 \cdot 2 \cdot 2=48$,
thus $\left(a_{5}+1\right)\left(a_{6}+1\right) \cdots \leqslant 3$.
So there is at most one $a_{j}(j \geqslant 5)$ that is a positive number, and $a_{j}$ can only be 1 or 2.
Consider the equation $\left(a_{1}+1\right)\left(a_{2}+1\right)\left(a_{3}+1\right)\left(a_{4}+1\right)\left(a_{5}+1\right)=144$.
For all solutions of $a_{1} \geqslant 3, a_{2} \geqslant 2, a_{3} \geqslant 1, a_{4} \geqslant 1, a_{5}=1$ or 2, and to make $n$ the smallest, we get $a_{1}=5, a_{2}=2, a_{3}=a_{4}=a_{5}=1$.
That is, $n=2^{5} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 11=110880$.