contest problem: number theory, prime factorization, perfect squares

Question

Problem Statement: Gretchen labels each of the six faces of a cube with a distinct positive integer so that for each vertex of the cube, the product of the three numbers on the faces touching the vertex is a perfect square. What is the least possible value of the sum of the numbers on this cube?

Ans: 55

Source: 2021 MathCounts

I'm thinking that we need each number to have a prime factorization with exponents of one. Two faces that share an edge will share that prime. But 2 faces that share an edge also share 2 vertices. So the 3 faces that share a vertex might be $2 \cdot 3$, $3\cdot5$, and $2 \cdot 5$. But then the face opposite of $2\cdot5$ would need to be at minimum $8\cdot 125$.

Or if we minimize the sum, by having two adjacent faces be $1$ and $2$, then the remaining faces could be odd powers of $2$.

I could use a hint or a solution. Thank you!

Answer 1

HINT: Put $2$ on the top face, $8$ on the bottom face. Then on the side faces, put $3,6,12,24$. So the resulting products on the top corners are respectively, $2 \times (3 \times 6)=6^2$, $2 \times (6 \times 12) = 12^2$, $2 \times (12 \times 24) = 24^2$, $2 \times (3 \times 24) = 12^2$. You can check likewise that the resulting products on the bottom products are all squares too.