# Thread: some help with number theory

1. ## some help with number theory

Show that if a positive integer $\displaystyle n=6a+5,a \geq 0$, then $\displaystyle n=3a'-1,a'\geq 1$.

2. Hello,
$\displaystyle n=6a+5=6(a+1)-6+5=6(a+1)-1=3\cdot2(a+1)-1$
$\displaystyle a' = 2(a+1)\ge 1$ because $\displaystyle a\ge 0$.

3. I do not know if this helps but:

I think that the key here is variable substitution.

$\displaystyle n=6a+5\Rightarrow$
$\displaystyle n=2\cdot 3\cdot a+2+3\Rightarrow$
$\displaystyle n=3(2a+1)+2$

By setting $\displaystyle b=2a+1$ we have:

$\displaystyle n=3b+2$

Next:
$\displaystyle n=3b+3-1\Rightarrow$

$\displaystyle n=3(b+1)-1\$

By setting $\displaystyle a'=b+1$ we have:
$\displaystyle n=3a'-1$

Think that this is that you want.
But you also check the proccess.

4. Hi

gdmath, you might want to check/edit your work. I think there's an error.

You got that $\displaystyle n=3b+2$, but the next line doesn't work out.

It says $\displaystyle n=3(b+1)+3-1$, but if you look at that, what we have is $\displaystyle n=3(b+1)+2=3b+5 \neq 3b+2$

I think what you should have is $\displaystyle n=3b+3-1=3(b+1)-1$

Then we set $\displaystyle a'=b+1$, leaving us with $\displaystyle n=3a'-1$

Also, $\displaystyle a'=b+1=2a+1+1=2a+2=2(a+1)$, which agrees with Taluivren's solution

Just one thing though ... shouldn't $\displaystyle a' \geq 2$?
Since when $\displaystyle a=0$, $\displaystyle a'=2(0+1)=2$