Show that if a positive integer $\displaystyle n=6a+5,a \geq 0$, then $\displaystyle n=3a'-1,a'\geq 1$.
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.
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 $\displaystyle a=0$, $\displaystyle a'=2(0+1)=2$