4. determine all natural numbers $n$ with exactly 100 different positive divisors, so that at least 10 of these divisors are consecutive numbers.

## Solution:
Of 10 consecutive integers, one is certainly divisible by 9, one by 8, one by 7 and one by 5. Therefore $n$ is divisible by $2^{3} \cdot 3^{2} \cdot 5 \cdot 7$ divisible. Now let

$$
n=2^{a} \cdot 3^{b} \cdot 5^{c} \cdot 7^{d} \cdot p_{1}^{e_{1}} \cdots p_{r}^{e_{r}}
$$

the prime factorization of $n$, where $r \geq 0, e_{k} \geq 1$ and $a \geq 3, b \geq 2, c, d \geq 1$. Then the number of positive divisors of $n$ is equal to

$$
(a+1)(b+1)(c+1)(d+1)\left(e_{1}+1\right) \cdots\left(e_{r}+1\right)
$$

By assumption, this is equal to $100=5 \cdot 5 \cdot 2 \cdot 2$, a comparison with the factors in the above formula shows that $a=b=4$ and $c=d=1$ as well as $r=0$ must apply. The only possible solution is therefore $n=2^{4} \cdot 3^{4} \cdot 5 \cdot 7=45360$. This number now actually has exactly 100 divisors, among which the 10 consecutive $1,2, \ldots, 10$ occur.