I have 2 numbers $n_1 = 1\pmod6$ and $n_5 = 5 \pmod 6$
I wish to know $\frac{4n_1-1}{3} \pmod 6$ and $\frac{2n_5-1}{3} \pmod 6$.
Since the answer varies depending on $n_1$ and $n_5$ I was wondering if there is some additional knowledge I might have about $n_1$ and $n_5$ that may pin down the answer. Perhaps their values mod 18?
Thanks in advance for any help and advice!
You know that $n_1=1+6k$, therefore $\frac{4n_1-1}{3}=1+8k\equiv 1+2k\bmod 6$.
Now $2k\bmod 6$ could be either $0,2$ or $4$ modulo $6$ depending on what $k$ is modulo $3$.
The knowledge of $k$ modulo $3$ is equivalent to the knowledge of $n_1$ modulo $18$.