Estonian Mathematical Olympiad

Since $100$ is divisible by $5^2$, but none of the numbers on the edges of the cube are divisible by $5^2$, the path of the first ant has to contain two numbers that are divisible by $5$. The only two such numbers are $5$ and $10$. Their product is divisible by $2$, but not by $2^2$. Since $100=2^2\cdot 5^2$, the third number on the path of the first ant has to be divisible by $2$ but not by $2^2$. The numbers on the edges of the cube that are divisible by $2$, but not by $2^2$, are $2$, $6$ and $10$. Since $10$ has already been used, the third number on the path of the first ant has to be either $2$ or $6$.

For the sum of the digits of the product of the second ant to be $2$, it has to consist of either one two or two ones, in addition to any number of zeroes. Since the product of the three largest numbers on the edges of the cube is $12\cdot 11\cdot 10=1320$, the possible products for the second ant are $2$, $11$, $20$, $101$, $110$, $200$, $1001$, $1010$ and $1100$. Of these $2$, $11$ and $101$ are prime numbers and cannot be the product of three integers that are larger than $1$. The number $20$ can also be eliminated, since the only way to write it as a product of three integers larger than one is $2\cdot 2\cdot 5$, which includes the number $2$ twice. The numbers $1001$ and $1010$ can be eliminated, since $1001=7\cdot 11\cdot 13$ and there is no $13$ on the edges of the cube, and $1010=2\cdot 5\cdot 101$ and there is no $101$ on the edges of the cube. This leaves $110$, $200$ and $1100$.

Now note that $200$ and $1100$ are both divisible by $5^2$. To get these products, there have to be two numbers on the path of the second ant that are divisible by $5$. The only such numbers are $5$ and $10$. Hence we can eliminate $1100$, since to get this product the third number would have to be $22$, which is not on the cube. To get the product $200$, the numbers should be $5$, $10$ and $4$. To get the product $110$ the numbers should be $2$, $5$ and $11$.

However, the paths of the two ants can have either $0$, $1$ or $3$ edges in common, but not exactly $2$ edges. Hence the numbers on the path of the second ant cannot be $5$, $10$ and $4$, and so the numbers on the path of the second ant have to be $2$, $5$ and $11$. This leaves the first ant with the numbers $5$, $6$ and $10$, which give a product of $300$.