2. (5 points) Two different natural numbers end with 7 zeros and have exactly 72 divisors. Find their sum.

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# Answer: 70000000

Solution: Since the number ends with 7 zeros, it has the form $N=10^{7} k$. The smallest number of this form $10^{7}$ has 64 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 7. 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, $10^{7} k=2^{a} 5^{b}$, and the number of its divisors is $(a+1)(b+1)$. The number 72 can be represented as a product of two factors greater than 7 in only one way: $72=8 \cdot 9$. Then, $N=2^{7} 5^{8}$ or $N=2^{8} 5^{7}$. Their sum is 70000000.