8. 1b.(TUR 5) Find the smallest positive integer $n$ such that (i) $n$ has exactly 144 distinct positive divisors, and (ii) there are ten consecutive integers among the positive divisors of $n$.
8. Among ten consecutive integers that divide $n$, there must exist numbers divisible by $2^{3}, 3^{2}, 5$, and 7. Thus the desired number has the form $n=$ $2^{\alpha_{1}} 3^{\alpha_{2}} 5^{\alpha_{3}} 7^{\alpha_{4}} 11^{\alpha_{5}} \cdots$, where $\alpha_{1} \geq 3, \alpha_{2} \geq 2, \alpha_{3} \geq 1, \alpha_{4} \geq 1$. Since $n$ has $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right) \cdots$ distinct factors, and $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+\right.$ $1)\left(\alpha_{4}+1\right) \geq 48$, we must have $\left(\alpha_{5}+1\right) \cdots \leq 3$. Hence at most one $\alpha_{j}$, $j>4$, is positive, and in the minimal $n$ this must be $\alpha_{5}$. Checking through the possible combinations satisfying $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{5}+1\right)=144$ one finds that the minimal $n$ is $2^{5} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 11=110880$.