Base 10 theorem in number theory

Question

The question asks:

Let $n\ \in \mathbb{N}$ in base $10$ as $n=a_ka_{k-1}\cdots a_1a_0$. Let $m=a_k+a_{k-1}+\cdots + a_1+a_0$. Then $9\mid n$ if and only if $9\mid m$

Here is my Proof:

$9\mid n$ implies that $n=9k$ where $k\in \mathbb{Z}$

Next, we can express $n$ as such: $$n=a_0+10a_1+100a_2+\cdots\\ n=a_0+(9+1)a_1+(99+1)a_2+\cdots\\ n=a_0+9a_1+a_1+99a_2+a_2+\cdots\\ n=(\underbrace{a_0+a_1+a_2+\cdots}_\text{m}) + 9(\underbrace{a_1+11a_2+\cdots}_\text{k})$$

Thus, $n=m+9k$ which implies that $9\mid(n-m)$ which can be written as the congruence, $n\equiv m\pmod 9$.

Now, from here, what steps do I need to take to prove the theorem?

You are done. You want to prove $n\equiv0\pmod9\iff m\equiv0\pmod9$ and you have proved $n\equiv m\pmod9$. — saulspatz, Apr 07 '21 at 23:09
So does $n\equiv m (\text{mod} , 9,)$ directly imply that $9|m$ and $9|n$? — Obsessive Integer, Apr 07 '21 at 23:12
No, it doesn't. $9|n$ is equivalent to $n\equiv0\pmod9$. The statement you have proved is stronger than the statement you set out to prove. — saulspatz, Apr 07 '21 at 23:14
Actually, I have one more question. Does this proof of mine hold for both sides of the if and only if statement? — Obsessive Integer, Apr 08 '21 at 05:52
Yes it does, if you state it that way. Say, "Thus $n=m+9k$ which is equivalent to $9\mid(n-m)$" instead of "which implies that" at the end. — saulspatz, Apr 08 '21 at 08:19

score -1 · Answer 1 · answered Apr 08 '21 at 00:46

-1

We have $k\in\Bbb Z$. Therefore:

If $9|n$ then $n/9\in\Bbb Z$ so $m/9=(n/9)-k\in\Bbb Z,$ so $9|m.$

If $9\not |\,n$ then $n/9\not\in\Bbb Z$ so $m/9=(n/9)-k\not\in\Bbb Z,$ so $9\not |\,m.$

answered Apr 08 '21 at 00:46

DanielWainfleet

57,985

Base 10 theorem in number theory

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