Pigeonhole principle and room full of flies

Question

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...

Hint: $5\times 27=135$, so you might consider 27 pigeonholes. — vadim123, Nov 07 '14 at 14:58
Equivalently show that of $136$ balls of radius $90\text{cm}$ in a cube of $3\text m$ side length at least $6$ have a common point. — AlexR, Nov 07 '14 at 14:58

VividD · Accepted Answer · 2014-11-07T15:14:34.060

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.

Pigeonhole principle and room full of flies

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