# contest problem: number theory, prime factorization, perfect squares

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!

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.
• How can you rule out labeling one of the faces with $1$? (Thank you for your reply!) Commented Apr 9, 2022 at 17:23
• If $1$ is a face, then a square has to be the opposite face, and then the remaining $4$ faces all need to either be (a) square or (b) nonsquare, such that any $2$ consecutive multiply to a square. I am trying to find a proof that isn't tedious, but for (b) essentially you run out of small numbers very quickly. For (a) all $6$ faces would be distinct squares which would sum to at least $1+4+9+16+25+36>55$.