3-4. In how many different ways can 1000000 be represented as a product of three natural ${ }^{1}$ numbers? Products that differ only in the order of the factors are considered identical.

(This problem was not solved by any of the olympiad participants.)
Solution 4. Let the factors have the form $2^{a_{1}} 5^{b_{1}}, 2^{a_{2}} 5^{b_{2}}$ and $2^{a_{3}} 5^{b_{3}}$. Then $a_{1}+a_{2}+a_{3}=6$ and $b_{1}+b_{2}+b_{3}=6$. Here, the numbers $a_{i}$ and $b_{i}$ can be zero. If $a_{1}=k$, then for the decomposition $a_{2}+a_{3}=6-k$ we get $7-k$ options. Therefore, for the decomposition $a_{1}+a_{2}+a_{3}=6$ we get $7+6+5+4+3+2+1=28$ options. In total, we get $(28)^{2}=784$ ways.

However, we have not yet taken into account the identity of decompositions that differ only in the order of the factors. There is exactly one decomposition, independent of the order of the factors, in which all factors are equal to 100. The decompositions in which there are two equal factors, we have counted three times. In each of the equal factors, 2 can enter in degrees $0,1,2$ or 3, i.e., in four different ways; 5 can also enter in four different ways. In total, we get 16 decompositions of this kind, but one of them is the decomposition considered above with three equal factors. There remain 15 decompositions, each of which we have counted three times. The number of decompositions with pairwise distinct factors is $784-1-45=738$. Each of them we have counted 6 times, so among them there will be $738 / 6=123$ different decompositions. In total, we get $1+15+123=139$ decompositions.