Prove $10^{n+1}+3\cdot 10^n+5$ is divisible by $9$? How do I prove that an integer of the form $10^{n+1}+3\cdot 10^{n}+5$ is divisible by $9$  for $n\geq 1$?I tried proving it by induction and could prove it for the Base case n=1. But  got stuck while proving the general case. Any help on this ? Thanks. 
 A: This can be proved by induction.
For $n=1$, the given expression becomes,
$10^{1+1}+3.10+5=135$ which is divisible by 9.
Let’s assume that the given statement is true for $k\ge1$, i.e., $10^{k+1}+3.10^k+5$ is divisible by $9$.
Then, $10^{(k+1)+1}+3.10^{k+1}+5=10^{k+2}+3.10^{k+1}+50-45$
=$10(10^{k+1}+3.10^k+5)-45$ is divisible by 9.
A: $10^{n+1}+3\cdot 10^{n}+5=10^{n}(10+3)+5=1300\cdots05$ has digit sum equal to $9$ and so is a multiple of $9$.
A: I would normally use congruences, but this can be done explicitly!
$$
10^{n+1}+3\cdot 10^{n}+5
= 9 \cdot \underbrace{11 \cdots 1}_{n+1} + 1
+
9 \cdot \underbrace{33 \cdots 3}_{n} + 3
+ 5
=\\=
9 \cdot (\underbrace{11 \cdots 1}_{n+1} + \underbrace{33 \cdots 3}_{n} + 1) = 9 \cdot 1\underbrace{44\cdots 4}_{n-1}5.
$$
A: Proof by induction:


*

*Base case: $10^{0+1}+3\cdot10^{0}+5=18$

*Assumption: $10^{n+1}+3\cdot10^{n}+5=9k$

*Inductive step:


$10^{n+2}+3\cdot10^{n+1}+5=$
$10^{n+2}+3\cdot10^{n+1}+50-45=$
$10(\color\red{10^{n+1}+3\cdot10^{n}+5})-45=$
$10(\color\red{9k})-45=$
$9(10k)-45=$
$9(10k-5)$
A: since $$10\equiv 1 \mod 9$$ we get
$$10^{n+1}+3\cdot 10^n+5\equiv 1+3+5=9\equiv 0\mod 9$$
A: Hint $\ {\rm mod}\ 9\!:\,\ \color{#c00}{10\equiv 1}\,\Rightarrow\,P(\color{#c00}{10})\equiv P(\color{#c00}1)\equiv $ sum of coefficients, for any polynomial $\,P(x)\,$ with integer coefficients, by the Polynomial Congruence Rule. See also Casting Out Nines.
A: Let $S(n)$ be the statement: $10^{n+1}+3\cdot{10^{n}}+5$ is divisible by $9$
Basis step: $S(1)$:
$\Rightarrow 10^{(1)+1}+3\cdot{10^{(1)}}+5=10^{2}+30+5$
$\hspace{45.5 mm}=100+35$
$\hspace{45.5 mm}=135$, which is divisible by $9$
Inductive step:
Assume $S(k)$ is true, i.e. assume that $10^{k+1}+3\cdot{10^{k}}+5$ is divisible by $9$
$\hspace{59 mm} \Rightarrow 10^{n+1}+3\cdot{10^{n}}+5=9A$
$\hspace{59 mm} \Rightarrow 10\cdot{10^{n}}+3\cdot{10^{n}}+5=9A$
$\hspace{59 mm} \Rightarrow 13\cdot{10^{n}}+5=9A$
$\hspace{59 mm} \Rightarrow 10^{n}=\dfrac{9A-5}{13}$
Then, $S(k+1)$: $10^{(k+1)+1}+3\cdot{10^{(k+1)}}+5$
$\hspace{23.5 mm} =10^{k+2}+3\cdot{10}\cdot{10^{k}}+5$
$\hspace{23.5 mm} =100\cdot{10^{k}}+30\cdot{10^{k}}+5$
$\hspace{23.5 mm} =130\cdot{10^{k}}+5$
$\hspace{23.5 mm} =130\cdot{\bigg(\dfrac{9A-5}{13}\bigg)}+5$
$\hspace{23.5 mm} =10\cdot{(9A-5)}+5$
$\hspace{23.5 mm} =90A-50+5$
$\hspace{23.5 mm} =90A-45$
$\hspace{23.5 mm} =9\hspace{1 mm}(10A-5)$, which is divisible by $9$
So, $S(k+1)$ is true whenever $S(k)$ is true.
Therefore, $10^{n+1}+3\cdot{10^{n}}+5$ is divisible by $9$.
