Pigeonhole principle and room full of flies

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Room is cube-shaped, with side lengths $3$ meters. $136$ flies flies are in it.

Prove that:

At any moment you can encompass $6$ flies with a sphere of radius $90$ centimeters.

This is from a math class, I couldn't devise an appropriate solution, but I guess it must be simple at the end of the day...

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Following up @vadim123 hint, it looks the following is the solution:

Divide the room in $27$ equal cubes (like Rubik's cube). At least one of these cubes contains $6$ flies (since $5*27=135<136$). And the sphere that contains all vertexes of such cube has radius $\sqrt3/2$ m which is less than $0.9$ m.


EDIT: It looks to me that the statement of the problem is too relaxed - in the sense that number 6 can be replaced with larger number. But the proof would look different of course.