Proof That 6 Divides Any Integer N

[Math100]

Homework Statement

Let $N=a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0}$, where $0\leq a_{k}\leq 9$, be the decimal expansion of a positive integer $N$.
Prove that $6$ divides $N$ if and only if $6$ divides the integer
$M=a_{0}+4a_{1}+4a_{2}+\dotsb +4a_{m}$.

Relevant Equations

None.

Proof:

Suppose that $6$ divides $N$.
Let $N=a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0}$, where $0\leq a_{k}\leq 9$, be the
decimal expansion of a positive integer $N$.
Note that $6=2\dotsb 3$.
This means $2\mid 6$ and $3\mid 6$.
Then $2\mid [4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m}]\implies 2\mid N\Leftrightarrow 2\mid [a_{0}+4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]$.
Now we have $3\mid (a_{0}+a_{1}+a_{2}+\dotsb +a_{m})\Leftrightarrow 3\mid [a_{0}+(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})+3(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]\Leftrightarrow 3\mid [a_{0}+4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]\Leftrightarrow 3\mid N$.
Observe that $6\mid N\Leftrightarrow 6\mid [a_{0}+4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]$.
Thus, $6$ divides the integer $M=a_{0}+4a_{1}+4a_{2}+\dotsb +4a_{m}$.
Conversely, suppose $6$ divides the integer $M=a_{0}+4a_{1}+4a_{2}+\dotsb +4a_{m}$.
Then $10^{m}\equiv 4\pmod {6}$ $\forall m\in\mathbb{N}\implies 10^{m}-4\equiv 0\pmod {6}$ $\forall m\in\mathbb{N}$.
This means $6\mid (10^{m}-4)$ $\forall m\in\mathbb{N}$.
Observe that $6\mid [(a_{0}+4a_{1}+\dotsb +4a_{m})+(10-4)a_{1}+(10^{2}-4)a_{2}+\dotsb +(10^{m}-4)a_{m}]\implies 6\mid (a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0})$.
Thus $6\mid N$.
Therefore, $6$ divides $N$ if and only if $6$ divides the integer $M=a_{0}+4a_{1}+4a_{2}+\dotsb +4a_{m}$.

 

[fresh_42]

Homework Statement:: Let $N=a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0}$, where $0\leq a_{k}\leq 9$, be the decimal expansion of a positive integer $N$.
Prove that $6$ divides $N$ if and only if $6$ divides the integer
$M=a_{0}+4a_{1}+4a_{2}+\dotsb +4a_{m}$.
Relevant Equations:: None.

Proof:

Suppose that $6$ divides $N$.
Let $N=a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0}$, where $0\leq a_{k}\leq 9$, be the
decimal expansion of a positive integer $N$.
Note that $6=2\dotsb 3$.
This means $2\mid 6$ and $3\mid 6$.
Then $2\mid [4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m}]\implies 2\mid N\Leftrightarrow 2\mid [a_{0}+4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]$.

I wouldn't use $\Leftrightarrow$ if you only prove one direction. You said $6\,|\,N.$
So $2 \,|\,N = 2\cdot 5 \cdot (a_{m}10^{m-1}+\dotsb +a_{2}10+a_{1}) + a_0$ and $2\,|\,a_0.$
Now, I see that $2\,|\,(a_{0}+4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})).$

Now we have $3\mid (a_{0}+a_{1}+a_{2}+\dotsb +a_{m})\Leftrightarrow 3\mid [a_{0}+(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})+3(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]\Leftrightarrow 3\mid [a_{0}+4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]\Leftrightarrow 3\mid N$.

Looks correct. But again, avoid the equivalence. We are only interested in one direction. It is easier to read as
\begin{align*}
3\mid (a_{0}+a_{1}+a_{2}+\dotsb +a_{m})&\Rightarrow 3\mid [a_{0}+(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})+3(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]\\
&\Rightarrow 3\mid [a_{0}+4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]
\end{align*}

Observe that $6\mid N\Leftrightarrow 6\mid [a_{0}+4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]$.
Thus, $6$ divides the integer $M=a_{0}+4a_{1}+4a_{2}+\dotsb +4a_{m}$.

The key is again that $2$ and $3$ are coprime. We have shown $6\,|\,N \Longrightarrow$
\begin{align*}
2&\mid [a_{0}+4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]\\
&\text{and}\\
3&\mid [a_{0}+4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]\\
&\Longrightarrow \\
2\cdot 3 = 6&\mid [a_{0}+4(a_{1}+a_{2}+a_{3}+\dotsb +a_{m})]
\end{align*}
But the last conclusion is only correct for coprime numbers.
E.g. $2\,|\,12 \wedge 4\,|\,12 \nRightarrow 2\cdot 4= 8\,|\,12.$ But $2$ and $3$ are coprime, so $2\,|\,12 \wedge 3\,|\,12 \nRightarrow 2\cdot 3= 6\,|\,12.$

Conversely, suppose $6$ divides the integer $M=a_{0}+4a_{1}+4a_{2}+\dotsb +4a_{m}$.

Then $10^{m}\equiv 4\pmod {6}$ $\forall m\in\mathbb{N}\implies 10^{m}-4\equiv 0\pmod {6}$ $\forall m\in\mathbb{N}$.
This means $6\mid (10^{m}-4)$ $\forall m\in\mathbb{N}$.
Observe that $6\mid [(a_{0}+4a_{1}+\dotsb +4a_{m})+(10-4)a_{1}+(10^{2}-4)a_{2}+\dotsb +(10^{m}-4)a_{m}]\implies 6\mid (a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0})$.
Thus $6\mid N$.
Therefore, $6$ divides $N$ if and only if $6$ divides the integer $M=a_{0}+4a_{1}+4a_{2}+\dotsb +4a_{m}$.

Correct. Only the layout isn't optimal:

Observe that
\begin{align*}
6&\mid [(a_{0}+4a_{1}+\dotsb +4a_{m})+(10-4)a_{1}+(10^{2}-4)a_{2}+\dotsb +(10^{m}-4)a_{m}]\\
&\implies \\
6&\mid (a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0}) =N
\end{align*}

 

[fresh_42]

If you want to prove both directions at once, you can use your second part and write:
\begin{align*}
6\,&|\,a_0+4(a_1+\ldots+a_m)=a_0+ 2\cdot (2a_1+\ldots+2a_m) = a_0+a_1+\ldots+a_m+3(a_1+\ldots+a_m)\\
&\Longleftrightarrow \\
2\,&|\,a_0 \wedge 3\,|\,(a_0+a_1+\ldots+a_m)\\
&\Longleftrightarrow \\
2\,&|\,N\wedge 3\,|\,N\\
&\Longleftrightarrow \\
6\,&|\,N
\end{align*}

 

[anuttarasammyak]

Interesting. Getting to know for m=1,2,3...

10^m =99..99+1 \equiv 1 (\mod 3)
10^m -4 \equiv 0 (\mod 3)
10^m -4 \equiv 0 (\mod 2)
So
10^m \equiv 4 (\mod 6)
a_m 10^m \equiv 4 a_m (\mod 6)
\sum_{m=1}^n a_m 10^m \equiv 4 \sum_{m=1}^n a_m (\mod 6)
a_0+\sum_{m=1}^n a_m 10^m \equiv a_0+4 \sum_{m=1}^n a_m (\mod 6)