I want to prove that $8\leq A_q(6,3)\leq 9$ (later on I'll have to prove $A_q(6,3)=8$ but for now they are just asking me to prove these bounds).
For the lower bound, I started off with a $Ham(r=3,q=2)$ code (which is a $[7,4,3]_2$-code and therefore a $(7,16,3)_2$-code), wrote down all of its words, took the ones that had a $0$ on the first letter and created a new code containing only these words, but getting rid of the first letter. This way I have a $(6,8,3)_2$-code and I have proved the lower bound.
However, I don't know how to deduce the upper bound. I cannot use Singleton's bound, and I can't think of anything else to do. Can someone help me?
Volume of Hamming sphere of radius $1$ is $(1+n)=7$ assuming $q=2.$ If $A_q(n,3)=10,$ this would imply that there are $10\times 7=70$ points in $\{0,1\}^6$ which is false (there are 64 points).