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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!

Ben Crossley
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1 Answers1

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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 $18$.

Arnaud Mortier
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