Given that $2n + 1$ is divisible by $3$, and $n$ is a natural number, what form can $n$ take? As per the solution I am referring to, $n$ must in the form $3k+1$, where $k$ is a whole number. The solution further adds that it is equivalent to $6p + 1$ or $6p + 4$.
While the answer makes sense when plugged into the expression $2n+1$, is there a generic set of steps to arrive at this answer? I probably seem to be missing something obvious :)
Thank you.
Odd multiple of $3$ can be written as $3(2k+1)=6k+3$
$$2n+1 = 6k+3$$
$$2n = 6k +2$$
$$n=3k+1$$
Exercise:
Explore what happens when $n=2p$.
Explore what happens when $n=2p+1$.
Consider $2n+1 = 3k$ modulo 2. Then $1 = k$ modulo 2, so $k = 2k'+1$, so $2n+1 = 3k = 6k' + 3$. So $2n+1 =6k' + 3$ which implies $2n = 6k' + 2$ or $n = 3k' + 1$.
so $n$ itself must be a multiple of 3 plus 1.
Also: all odd numbers that are multiples of 3, are of the form multiples of 6 plus 3.
While there are different ways of approaching this, as described in other two answers, I found Remainder theorem approach easy to follow.
Given that $2n + 1$ is divisible by 3,
$Rem (\frac{2n + 1}{3}) = 0$ $=>$ $Rem (\frac{2}{3})*Rem (\frac{n}{3}) + Rem (\frac{1}{3}) = 0$. solving this, we get $Rem (\frac{n}{3}) = 1$. Hence $n$ can take the form $3k + 1$.
Let $n=3m+r$, with $0\le r<3$; then $$ 2n+1=6m+2r+1\equiv2r+1\pmod 3 $$ Thus we have $2r+1\equiv 0\pmod{3}$. Therefore $r\equiv 1\pmod{3}$ (prove it). Also the converse holds (prove it).
In order to find the other form of the answer, distinguish between $m$ even or odd.
