2. (5 points) Two different natural numbers end with 8 zeros and have exactly 90 divisors. Find their sum.
Answer: 700000000

Solution: Since the number ends with 8 zeros, it has the form $N=10^{8} k$. The smallest number of this form $10^{8}$ has 81 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 8. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ has other prime divisors, the number of divisors of $N$ increases by at least a factor of two: in addition to the original $1,2,5,10,20 \ldots$, there will also be $k, 2 k, 5 k, 10 k, 20 k$ and so on. Therefore, $10^{8} k=2^{a} 5^{b}$, and the number of its divisors is $(a+1)(b+1)$. The number 90 can be represented as a product of two factors greater than 8 in only one way: $90=9 \cdot 10$. Then, $N=2^{8} 5^{9}$ or $N=2^{9} 5^{8}$. Their sum is 700000000.