find the maximum value of $m$ where $3^m$ be a factor of $A_{1395}$. 
Consider the number $A_{1395}=\underbrace{333\cdots3}_{1395}$. what is
the maximum value of $m$ where $3^m$ be a factor of $A_{1395}$.
$a) 2\quad\quad b)3\quad\quad c)4\quad\quad d)5\quad\quad e)6$

Here is my attempt :
We have $A_{1395}=3\times \underbrace{111\dots1}_{1395}$ sum of the digits of $\underbrace{111\cdots1}_{1395}$ is $1395$ which is a multiply of $9$. therefor this number is dividable to $9$. so we conclude $A_{1395}$ has the factor $3^3$ . to see whether it has more factor of $3$ or not I divided it to $27$. after doing some steps of long division I got:
$$\underbrace{11\cdots1}_{1395}÷27=4115226\cdots$$
Unfortunately I couldn't see any pattern as the result. so I don't know how to check this number has more factor of $3$ or not.
 A: Write using the binomial theorem,
\begin{align}
999\cdots 9 &= 10^{1395} -1 \\
&= (1+9)^{1395} -1 \\
&= \left(\array{1395\\1}\right)\cdot 9 + \left(\array{1395\\2}\right)\cdot9^2 + \cdots 9^{1395}
\end{align}
But $1395 = 3^2 \cdot 5 \cdot 31$.  So, the first term is divisible by $3^4$ and each term thereafter is divisible by $3^5$ meaning the sum is divisible $3^4$ and not by $3^5$.  Accordingly, $333\cdots 3$ is divisible by $3^3$ but not $3^4$.  The answer is therefore $3$.
A: We write the number as ${10^{1395}-1\over 3}$
By Euler Thorem,
$10^{162}\equiv1\pmod{243}$ so $10^{1395}\equiv10^{99}\pmod{243}$
Since $10^{99} \equiv 28^{33}\equiv82^{11}\equiv(-80)^5\times82\equiv82^2\times(-80)\times82\equiv(-80)^2\times82\equiv-80\equiv163\pmod{243}$
We know $10^{99}-1\equiv162\pmod{243}$ which implies $10^{1395}-1\equiv 162\pmod{243}$
Therefore $10^{1395}-1$ is not a multiple of $243$ but is a multiple of $81$.
So ${10^{1395}-1\over 3}$ is a not multiple of $81$ but is a multiple of $27$
A: For $x\in\Bbb N$, letr $v_3(x)$ be the exponent of the maximal power of $3$ dividing $x$.
If $n=3^ka+1$ with $k\ge1$ and $3\nmid a$, then
$$ n^3=3^{3k}a^3+3^{2k+1}a^2+3^{k+1}a+1
=3^{k+1}(a+3^k(\ldots))+1$$
i.e.,

If $v_3(n-1)>0$ then $v_3(n^3-1)=1+v_3(n-1)$.

Note that $A_{1395}=\frac{10^{1395}-1}3$, so what we are looking for is $m=v_3(10^{1395}-1)-1$.
By the above remark,
$$m=v_3(10^{1395}-1)-1=v_3(10^{465}-1) = v_3(10^{155}-1)+1. $$
Also, as $v_3(10^3-1)=1+v_3(10-1)=3$, so that
$$ 10^{155}=(10^3)^{51}\cdot 100\equiv 100\equiv 19\pmod{27}$$
(but $\equiv 1\pmod 9$). We conclude
$$v_3(10^{155}-1)=2 $$ and ultimately
$$ v_3(A_{1395})=3.$$
A: Consider an extension 9f the familiar sum of digits test for divisibility by $9$.
The number $999$ factors as $27×37$ or with prime factors, $3^3×37$.  Thus if you add three-digit blocks together to cast out multiples if $999$, you are also casting out multiples of $27$ and $37$ and thus the sum of tyree-digit blocks tests for divisibility by these factors.
This can be extended to higher powers of $3$.  The sum of nine-digit blocks tests for divisibility by $81$, and more generally adding blocks of $3^k$ digits tests for divisibility by $3^{k+2}$
In your case, then, first sum the three-digit blocks to get  $333×465$ where each of $465$ blocks is $333$.  Since $333$ is a multiple of $9$ and $465$ is a multiple of $3$, your three-digit block sum passes for divisibility by $27$.
Next try the nine-digit block sum getting $333333333×155$.  The three-digit blocks of $333333333$ add up to a multiple of $27$, but dividing out a factor of $3$ gives $111111111$ which fails that test.  So $333333333$ is a multiple of $27$ but not $81$, and $155$ is not a multiple of $3$, so the nine-digit block sum which equals their product fails divisibility by $81$.
Thus the maximum power of $3$ that divides $A_{1395}$ is $3^3=27$.
