Prove that $(\forall x \in \mathbb{N})(\exists y \in \mathbb{N})(\underbrace{3...3}_\text{x}|\underbrace{1...1}_\text{y})$ and find the smallest $y$ 
Prove that $\left(\forall x \in \mathbb{N}\right)\left(\exists y \in
\>\mathbb{N}\right)\left(\underbrace{3...3}_\text{x}|\underbrace{1...1}_\text{y}\right)$
  and find the smallest $y$ for each $x$.
Note: "$a|b$" means "$a$ divides $b$".

I know that for $x \in \mathbb{N}, k \in \mathbb{Z}:$
$$3|\underbrace{111}_\text{3}, 3|\underbrace{111111}_\text{6}, 3|\underbrace{1...1}_\text{9}, 3|\underbrace{1...1}_\text{3k};
\\
33|\underbrace{111111}_\text{6}, 33|\underbrace{1...1}_\text{12}, 33|\underbrace{1...1}_\text{$3 \cdot 2 \cdot k$};
\\
\underbrace{3...3}_\text{x}|\underbrace{1...1}_\text{$3 \cdot x \cdot k$}.$$
So $y=3xk$ satisfies the conditions and the smallest $y$ equals $3x$, but how can i prove this formally?
My attempt at solution:
Edit:
I wanted to prove that $y$ must be $3lx$ and so solve both problems. But now I realized that my method does not work. Now I want to understand how to prove that $3x$ is the smallest possible y if I prove (by induction or by any other method) that there always exists $k$ for $y = 3lx$.
$$\left(\forall x \in \mathbb{N}\right)\left(\exists y \in \mathbb{N}\right)\left(\underbrace{3...3}_\text{x}|\underbrace{1...1}_\text{y}\right)
\\
\iff 
\\
\left(\forall x \in \mathbb{N}\right)\left(\exists y \in \mathbb{N}\right)\left(\left(\exists k \in \mathbb{Z}\right)\left(\sum_{i=0}^{y-1} 10^i = 3 \cdot \sum_{i=0}^{x-1} 10^i \cdot k\right)\right)$$
By the definition of repunit (number that contains only the digit 1):$$
\sum_{i=0}^{n-1} 10^i = \dfrac{10^n-1}{9}, n \geq 1,$$
then
$$\dfrac{10^y-1}{9} = 3k\cdot\dfrac{10^x-1}{9}.$$
So $\dfrac{10^y-1}{9}$ is divisible by 3 and $\dfrac{10^x-1}{9}$.
For 3:
$$\dfrac{10^y-1}{9} = 3l
\\
\underbrace{1...1}_\text{$y$} = 3l
\\
\underbrace{1 +\ ...\ + 1}_\text{$y$} = 3l\ (*)
\\
y = 3l$$
(*): Divisibility rule for 3.
For $\dfrac{10^x-1}{9}$:
$$\dfrac{10^y-1}{9} = \dfrac{10^x-1}{9} \cdot m
\\
y = xm\ (**)$$
(**): Proof: Prove that if $\frac {10^n-1}{9} | \frac {10^m-1}{9}$, then $n|m$.
So $y = lcm(3, x) = \dfrac{3x}{gcd(3,x)}$ and i stuck there. Any ideas please?
Thanks in advance.
 A: You more or less have it. 
Let $333......33$ $k$ threes and $11111.....1$ $k$ times be written as $3_k$ and $1_k$ respectively.
$3_x = 3*1_x|1_y$ will mean $3|1_y$ $1_y$ has a multiple of $3$ $1$s.  It will also mean $1_x|1_y$ which, as you researched, means $x|y$.  
So if $3\not \mid y$ then $x|y \implies 3\not \mid x$ we need $y$ to be the smallest multiple of both $x$ and $3$.  That is ... $y = 3x$.
But what if $3|x$? And so $x|y$ and $3|y$. Say $x=3^mk$ and $\gcd(3,k) = 1$  Then $1_x = 1_k*100....10000....1:= 1_k*1_{3^m,k}$ where $1_{a,b} =$ $a$ $1$ separated by $b-1$ $0$s (or other words $\sum_{j=0}^{a} 10^{j*b}$).
So $3_x|1_y$ means $3|1_y$ and $1_x|1_y$ and $1_k|1_y$ and $1_{3^m,k}|1_y$. 
Let $V = \frac {1_y}{1_3,k}$ which must be equal to $1_j$ where $k|j$.  But then $3*1_k|V$ so $V$ is divisible by $3$ and $3|j$.  So $y$ must be a multiple of $3x$.
So the smallest such multiple of $x$ satisfiying all that  is still $3x$.
A: The existence proof is easy.  Suppose there are $n$ 3's in your proposed divisor $d$.  Form the number $M$ with $3n$ 1's.  Then since the number of 1's in $M$ is divisible by $n$, $M$ will be divisible by $d/3$ (a number with $n$ 1's).  By the digital sum rules for divisibility by powers of $3$ (for divisibility by $3^k$ with $k\ge 2$, blocks of $3^{k-2}$ digits must add up to a multiple of $3^k$), tripling the number of 1's in $M$ from $n$ to $3n$ will then provide an additional factor of $3$ that completes the divisibility requirement.
Example:  $n=3, d=333$ gives $111,111,111$ as the candidate.  We then have $111,111,111=1,001,001×111=333,667×333$.
This does NOT, however prove that smallest solution.
A: $33...33 = 3 ˟ 11...11 $
if number of digits of $11...11 $ is $2p$,(p=1,2,3...) then it is divisible by $11...11$ with number of digits $2q≤ p$, i.e. we must have:
$n_1=2p$
Also $11...11$ is divisible by $3$ if the sum of digits is divisible by 3. That means we must have:
$n_1 = 3 k$
That is if the number of digits of $11...11$ is $n_1=6pk$ or generally $n_1=6 t$ (t any natural number) then a number like $33...33$ can divide it if the number of its digits is $n_3=6s, n_3≤ n_1/2$
