Prove $10^{3n+1}+10^{6n+2}+1=111k$ I'm stuck to prove this one can anyone help me please?

Prove that $10^{3n+1}+10^{6n+2}+1=111k$.

I'm not sure what exactly I should do.
n is a natural number and k is an integer
 A: The OP has corrected the problem.  The corrected answer is below.
Assuming that $k$ must be an integer (not stated by OP), this is the same as showing that
$$
10^{n+1}+10^{6n+2}+1\equiv 0\pmod{111}.
$$
Observe that $10^3=1000=999+1=1+9(111)$.  Therefore, there are only three values for the first and second terms, $10$, $100$, and $1$.  In particular, when
$$
10^{n+1}=\begin{cases}10&n+1\equiv 1\pmod 3\\100&n+1\equiv 2\pmod 3\\1&n+1\equiv 0\pmod 3.\end{cases}=\begin{cases}10&n\equiv 0\pmod 3\\100&n\equiv 1\pmod 3\\1&n\equiv 2\pmod 3.\end{cases}
$$
On the other hand, there is only one value for $10^{6n+2}\pmod{111}$ since $10^{6n}=(10^3)^{2n}\equiv 1^{2n}=1\pmod{111}$.  Therefore, the second term is equivalent to $100$ modulo $111$.
This appears to be a problem, since (modulo $111$) there are exactly three sums $10+11+1$, $100+100+1$, and $1+100+1$, only the first one is equivalent to $0$ modulo $111$.  In other words, if we let $n=1$, then the sum is $100,000,101$, which is not a multiple of $111$.
To the OP, please check the original statement.  The statement would be correct if the first term were $10^{3n+1}$, however.
When the first term is $10^{3n+1}$ instead of $10^{n+1}$, observe that $10^{3n}=(10^3)^n\equiv 1^n=1\pmod{111}$, therefore, the sum can be reduced to
$$
10^{3n+1}+10^{6n+2}+1\equiv 10+10^2+1\equiv 0\pmod{111}.
$$
A: There's a typo. The assertion should be
$$10^{3n+1}+10^{6n+2}+1\equiv 0\mod111.$$
As  $10^3\equiv1 \bmod 3$ and$\bmod37$, and $3$ and $37$ are coprime, $10^3\equiv 1\bmod 111$. Hence
$$10^{3n+1}+10^{6n+2}+1\equiv 10+10^2+1=111\equiv0\mod111.$$
A: Hint $\ {\rm mod}\,\ i^2\!+i+1\!:\,\ \color{#0a0}{i^3}\equiv\color{#c00}{\bf 1}\ $ by $\ i^3\!-1 = (i\!-\!1)(i^2\!+i+1)\equiv 0\ $
therefore  $\qquad\quad  \begin{align} &\color{#0a0}i^{\large \color{#0a0}{3}j}\, i^{\large  2} + {\color{#0a0}i^{\large\color{#0a0}{3}k}}\, i +  1\\ 
\equiv\ &\,\color{#c00}{\bf 1}^{\large  j}\,i^{\large 2}\,  + \color{#c00}{\bf 1}^{\large  k}\, i + 1\\
\equiv\ &\quad\ i^{\large 2}\ \  +\,\ \ \ i+1\ \equiv\ 0 \end{align}\qquad $ OP is case $\,i = 10,\ j=2n,\ k=n$
