# Thread: some help with number theory

1. ## some help with number theory

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

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

3. I do not know if this helps but:

I think that the key here is variable substitution.

$
n=6a+5\Rightarrow
$

$
n=2\cdot 3\cdot a+2+3\Rightarrow
$

$
n=3(2a+1)+2
$

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

$
n=3b+2
$

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

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

By setting $a'=b+1$ we have:
$
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 $n=3b+2$, but the next line doesn't work out.

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

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

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

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

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