Let the vertices of the graph G be the numbers 1, 2, ..., 100, and two (different) vertices be adjacent if and only if at least one of 2, 3, or 5 is a common divisor of the respective numbers. Determine χ(G), the chromatic number of the graph G.

I tried to solve and got that all even numbers i.e 2,4,6,...,100 form a clique of size 50. And that all prime numbers are isolated so they do not matter in coloring. But I cannot show chromatic number is 50. Any suggestions?

  • $\begingroup$ All even numbers form a clique, as they have $2$ as a common divisor. You will need at least $51$ colors. $\endgroup$
    – Eric_
    Jun 9 at 16:29
  • $\begingroup$ HOW? Number of even numbers are 50 not 51 $\endgroup$ Jun 9 at 16:33
  • $\begingroup$ In addition to the 25 prime numbers, there are 4 other isolated vertices: 1, 49, 77, 91. $\endgroup$
    – Stef
    Jun 9 at 21:33

1 Answer 1


Hint 1:

Show that the graph is $50$-partite.

Hint 2:

Each partite set contains exactly two elements.

Full Solution:

For colors $c_1,\dotsc,c_{50}$, assign color $c_i$ to numbers $2i-1$ and $2i$. Note that no two consecutive numbers can have a common divisor other than $1$.

  • 1
    $\begingroup$ Very smart. Thank you $\endgroup$ Jun 9 at 17:23

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