What's the relation that joins the nodes? Open the image in a new tab if you'd like to see the diagram with better resolution.

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Hint 1

It is equally important to think about why any given two nodes are connected as it is to consider why they are not connected.


2 Answers 2


The numbers are connected if the prime representation of them has the same number of primes (can be non-distinct)


$24=2^3\times3^1$ so there are $3+1=4$ prime factors and $16=2^4$ which has $4$ prime factors so they are connected.
$11=11^1$ so there are $1$ prime factor and $12=2^2\times3^1$ so there are $2+1=3$ prime factors so they are not connected.

I thought it should be nice to include my thinking process:

All the primes are connected to each other. What is the same among the primes?
Oh! They have the same number of factors($2$)!
No, it doesn't applies for some connections.
Each graph is a complete graph.
After a while, I found out the connection between the numbers.

  • $\begingroup$ Thanks for sharing your thought process! It's cool to see that you overcame the less obvious part of this puzzle. $\endgroup$
    – Galen
    Apr 29, 2020 at 5:06

Solving process:

The graph clearly has four connected components, one of which contains all the primes from 2 to 19. In attempting to see what was connected to what within these components, and indeed if anything was disconnected, I found that in fact each connected component was complete -- i.e. the original graph was a labelling of $K_9 \cup K_8 \cup K_4 \cup K_2$. Thus, since the graph was undirected and each connected component obeyed transitivity, the adjacency relation was some kind of equivalence relation. So since one such equivalence class was "all the primes," I checked the factorizations of the rest of the nodes, and each connected component had nodes with the same length of factorization (without exponents, i.e. $2*2*2*3$ instead of $2^3*3$). Therefore...


The adjacency relation is "has the same number of prime factors, including multiplicity."

  • $\begingroup$ This answer appears to be a duplicate of the already accepted answer. $\endgroup$
    – fljx
    Jan 28, 2022 at 23:02

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