Argentina_2018

Let the numbers on the pair of opposite faces be $a_1,a_2,b_1,b_2,c_1,c_2$. Note that $a_1$ participates in 4 products: $a_1{b}_1{c}_1,a_1{b}_1{c}_2,a_1{b}_2{c}_1,a_1{b}_2{c}_2$. Likewise $a_2$ participates in the remaining 4 products and in a completely analogous fashion the products are $a_2{b}_1{c}_1,a_2{b}_1{c}_2,a_2{b}_2{c}_1,a_2{b}_2{c}_2$. It follows that the 8 products have sum

$$a_1(b_1{c}_1+b_1{c}_2+b_2{c}_1+b_2{c}_2)+a_2(b_1{c}_1+b_1{c}_2+b_2{c}_1+b_2{c}_2)=(a_1+a_2)(b_1+b_2)(c_1+c_2).$$

Then by hypothesis $(a_1+a_2)(b_1+b_2)(c_1+c_2)=315$. Note that each factor on the left is greater than 1. Thus $a_1+a_2,b_1+b_2,c_1+c_2$ are divisors of 315 that are greater than 1 and have product 315. Conversely, let $d_1,d_2,d_3$ be divisors of 315 with these properties. Since $d_i>1$, one can write $d_1=a_1+a_2,d_2=b_1+b_2,d_3=c_1+c_2$ with $a_1,a_2,b_1,b_2,c_1,c_2$ natural numbers. Place them on the faces of the cube (so that the $a_i$'s are on opposite faces; the same for the $b_i$'s and the $c_i$'s). Then, by the above, the sum of the 8 vertex products $a_i{b}_j{c}_k$ equals $(a_1+a_2)(b_1+b_2)(c_1+c_2)=d_1{d}_2{d}_3=315$.

Note that the representations $d_1=a_1+a_2,d_2=b_1+b_2,d_3=c_1+c_2$ can be chosen in different ways. But the sum $a_1+a_2+b_1+b_2+c_1+c_2$ which we are interested in is the same, equal to $d_1+d_2+d_3$ and depending only on the triple of divisors $d_1,d_2,d_3$ with the properties stated above. In this way the question reduces to finding all factorizations $315=d_1{d}_2{d}_3$ with factors $d_i>1$. All solutions to the problem are the respective sums $D=d_1+d_2+d_3$. Since $315=3^2\cdot 5\cdot 7$, consider two cases. If one of the $d_i$'s be divisible by $3^2=9$ then it is immediate that the factorization is $5\cdot 7\cdot 9$ and $D=5+7+9=21$. Otherwise two $d_i$'s are exactly divisible by 3, and it is clear that one of them is in fact equal to 3. It remains to factorize $3\cdot 5\cdot 7$ into two factors greater than 1, which can be done in 3 ways: $15\cdot 7\cdot 3$, $35\cdot 3\cdot 5$, $21$. Hence the remaining possibilities for admissible factorizations of 315 are $3\cdot 15\cdot 7$, $3\cdot 35\cdot 3$, $5\cdot 21$. They yield respectively $D=25,41,29$. In summary the answer to the problem is 21, 25, 29, and 41.