2. (5 points) Two different natural numbers end with 9 zeros and have exactly 110 divisors. Find their sum.
Answer: 7000000000

Solution: Since the number ends with 9 zeros, it has the form $N=10^{9} k$. The smallest number of this form $10^{9}$ has 100 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 9. 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, $10^{9} k=2^{a} 5^{b}$, and the number of its divisors is $(a+1)(b+1)$. The number 110 can be represented as a product of two factors greater than 9 in only one way: $110=10 \cdot 11$. Thus, $N=2^{9} 5^{1} 0$ or $N=2^{1} 05^{9}$. Their sum is 7000000000.