2. (5 points) Two different natural numbers end with 6 zeros and have exactly 56 divisors. Find their sum.
Answer: 7000000

Solution: Since the number ends with 6 zeros, it has the form $N=10^{6} k$. The smallest number of this form $10^{6}$ has 49 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 6. 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, $10^{6} k=2^{a} 5^{b}$, and the number of its divisors is $(a+1)(b+1)$. The number 56 can be represented as a product of two factors greater than 6 in only one way: $56=7 \cdot 8$. Then, $N=2^{6} 5^{7}$ or $N=2^{7} 5^{6}$. Their sum is 7000000.