Show that there is no q-ary code of length 5 and distance 3 with more than $\displaystyle \frac{q^5}{5q-4}$ codewords. You should prove this from first principles using spheres and without using known bounds on the number of codewords.

This is my working so far:

First prove that there is a q-ary code of length 5 and distance 3 with exactly $\displaystyle \frac{q^5}{5q-4}$ codewords. consider the spheres of radius 1 centred on codewords. Each such sphere contains 1 + 5 = 6 words. Since the code has distance 3, no two such spheres intersect. Otehrwise we would have a pair of codewords at distance 2 or less. Since there are $\displaystyle \frac{q^5}{5q-4}$ codewords, there are $\displaystyle \frac{6q^5}{5q-4}$ distinct words in these spheres. However there are only $\displaystyle q^5$ distince q-ary words of length 5. therefore we have to show that

$\displaystyle \frac{6q^5}{5q-4}$ = $\displaystyle q^5$ for some q that is a natural non-zero number.

i worked out q = 2 therefore such a code exists. but now im having problems proving that no q-ary code with MORE THAN $\displaystyle \frac{q^5}{5q-4}$ codewords exists. also, with q = 2 the no of codewords = $\displaystyle \frac{16}{3}$ can anyone tell me why this isnt a whole number?

PLEASE HELP!!! any help or advice would be GREATLY appreciated!!!