planar graph of Submultiples [duplicate]

There is Graph which is connected with Submultiples. (I am sorry but I don't know what this is called.)

For example,

10-node Graph has 10 nodes, 18 edges. node 1 connect all the other nodes. node 2 connect node number 2,4,6,8,10. node 3 connect 3,6,9. node 4 connect 4,8. etc. Image of Planar graph of 10-node Submultiple graph This is Planar graph of 10-node graph.

Question 1.

It is easy to check if 11-node graph is planar or not. (Just put the node no.11 and connect with no.1) But, Is 12-node Submultiple graph planar? And how can I prove it? (I saw Kuratowski's theorem But I cannot find the graph of $$K_{3,3}$$ If there is, please show me with picture.

Question 2.

If 12-node Submultiple graph is possible, How many node are possible? (I mean, 14 or 15 nodes graph is planar? or not?)

marked as duplicate by saulspatz, Shailesh, Lord Shark the Unknown, YuiTo Cheng, CesareoMay 27 at 7:43

• I think the word you're looking for is "divisors." – saulspatz May 26 at 21:45

Here is a planar graph for $$12$$ nodes:
The thicker red edges are the onesthat I changed their orientations. Green edges are the ones that are incident to node $$12$$.
From here, I can say that for $$13$$ and $$14$$ nodes, it can be still planar. $$13$$ is similar to $$11$$ since $$13$$th node is just adjacent to $$1$$ and for $$14$$, we can add $$14$$ to outer face in which $$2$$ and $$7$$ reside.
For $$15$$, it is not trivial. But there is a $$K_{3,3}$$ subgraph using subdivions as the following:
For $$16$$ nodes, I am sure that it cannot be planar because the nodes $$1,2,4,8,16$$ will all be connected to each other so we have $$K_5$$ as a subgraph. Then by Kuratowski's Theorem, it is not planar.