For all $x\in\mathbb{N}^*$ there is some $y\in\mathbb{N}^*$ such that $xy$ contains only $0$s and $1$s. The following is an interesting problem a friend of mine told me.

Let $\mathbb{N}^*=\mathbb{N}\setminus\{0\}$ and $U\subset \mathbb{N}^*$ be the set of positive integers whose base $10$ representation contains only $0$s and $1$s.
Show that for all $x\in\mathbb{N}^*$ there is some $y\in\mathbb{N}^*$ with $xy\in U$.

I have managed to solve it, but the solution is rather inelegant. Moreover, it uses Euler's theorem (on the totient function), and my friend implied that there should be a 'basic solution' -- that this problem should be solvable by an undergrad in an introductory course to real analysis.
I am interested to see what different solutions people can come up with. I can also post my own answer, but I figured I'd let you guys try your own hand at it before introducing some 'bias'.
 A: Consider the following sequence
$$a_i = \sum_{j=0}^i 10^j=\overbrace{1\ldots1}^{i+1}$$
where $i$ takes value from $0$ to $x$.
That is $a_0=1, a_1=11$,$a_2=111$, and so on, where $a_x=\overbrace{1\ldots1}^{x+1}$.
By pigeonhole principle, when we take modulo $x$, two of them will share the same remainder, $r$. Let these two indices be $i$ and $j$, $t<s$, hence $a_t < a_s$.
$$a_t = xp + r$$
$$a_s=xq+r$$
$$a_s-a_t=x(q-p)$$
Notice that $a_s-a_t$ consists of only $0$ and $1$ and it is positive.
That is $a_s-a_t \in U$ and we can pick $y=q-p$.
A: Well, since Siong very quickly provided an awesome answer, I will show my own (rather shameful) answer for those that are curious.

Fix $x \in\mathbb{N}^*$. Notice that we may assume without loss of generality that $(x,10)=1$.
Let $X=x\cdot\mathbb{N}^*$ and $U_0=U$.
Observe that the problem is equivalent to showing that $X\cap U_0$ is nonempty.
Choose some $k_1,n_1\in\mathbb{N}$ with the property that $xn_1+k_1=u_0\in U_0$.
It's easy to see that one can always pick $k_1<x$, although it's not necessary.
Now, let $U_1\subset\mathbb{N}^*$ be the set obtained by adjoining $k_1$ to the left of each $u\in U$, in base $10$ notation.
Suppose that $X\cap U_1$ is nonempty, so that for some $a\in\mathbb{N}$ one may write
$$x\cdot a=k_1\,\square$$
where $k_1\,\square$ is a number whose base $10$ representation has $k_1$ to the left, and $\square$ denotes the rest of its base $10$ representation, which consists solely of $0$s and $1$s.
If there are $d$ digits in $\square$, we may write $k_1\,\square=k_1\cdot 10^d+v$ for some $v<10^d$ in $U$.
In this case:
\begin{align}
x\cdot (a+n_1\cdot10^d)
&=xa+xn_1\cdot10^d\\
&=\left(k_1\cdot10^d+v\right)+xn_1\cdot10^d\\
&=(xn_1+k_1)\cdot10^d+v\\
&=u_0\cdot10^d+v
\end{align}
In other words, $x\cdot (a+n_1\cdot10^d) \in U_0$.
This means that the problem is equivalent to showing that $X\cap (U_0\cup U_1)$ is nonempty.

Now, suppose that we do not yet know whether $X\cap U_1$ is nonempty.
We can repeat the process as follows.
Choose $k_2,n_2\in\mathbb{N}$ with the property that $xn_2+k_2=u_1\in U_1$ and let $U_2\subset\mathbb{N}^*$ be the set obtained by adjoining $k_2$ to the left of each $u\in U$, in base $10$ notation.
If $X\cap U_2$ is nonempty, one may write $x\cdot a =k_2\cdot 10^d+v$ for some $a,d\in\mathbb{N}$ and $v<10^d$ in $U$.
Proceeding as above, we will find that $x\cdot(a+n_2\cdot 10^d)=u_1\cdot10^d+v\in U_1$, which means that, like before, the problem is equivalent to showing that $X\cap(U_0\cup U_1 \cup U_2)$ is nonempty.

At this point, I believe the induction is clear.

Lemma: For each $j\geq1$, we may choose $k_j$ so that $k_j\equiv jk_1\pmod x$.

The lemma show that in at most $x$ steps we will have $u_{x-1}=xn_x+k_x\equiv 0\pmod x$, that is, $u_{x-1}\in X\cap U_{x-1}$, which completes the proof.
Proof of Lemma:
Since $x$ and $10$ are coprime, by Euler's theorem we have that
$$10^{\varphi(x)}\equiv 1 \pmod x.$$
In particular, there are arbitrarily high powers of $10$ equivalent to $1$ modulo $x$.
We define the $k_j$ via induction.
The base case $j=1$ is obvious, and remember it entails the existence of $n_1\in\mathbb{N}$ and $u_0\in U_0$ with $xn_1+k_1=u_0$.
For the inductive step, suppose $k_j$ has been defined so that $k_j\equiv jk_1\pmod x$.
Choose some $d\in\mathbb{N}$ with $10^d\equiv 1 \pmod x$ and $u_0<10^d$.
Then:
$$k_j\cdot 10^d+u_0=k_j\cdot 10^d+xn_1+k_1=xn_1+\left(k_j\cdot 10^d+k_1\right).$$
Notice that $\left(k_j\cdot 10^d+u_0\right)\in U_{j}$ and that $k_j\cdot 10^d+k_1\equiv k_j+k_1\equiv (j+1)k_1\pmod x$.
It follows that we may take $n_{j+1}=n_1$ and $k_{j+1}=k_j\cdot 10^d+k_1$. $\square$
A: This is only a part of the answer. Consider this as a hint. 
Basically, after you choose $x$, what you have to do is, look at the decimal expansion of $\frac 1x$. Since ${x}$ is natural, ${\frac 1x} $ will be either a terminating or recurring decimal. If it is terminating, then $x$ must be a multiplication of powers of 2 and 5. Then you are done.
Now consider $x$ to be a prime other than 2, 3, 5. Then $\frac 1x$ will be a recurring decimal.
 For example take $x=7$. Then $$1/7=0.\bar{abcdefg} =\frac {abcdefg} {999999}$$ So $7×abcdefg=999999$.
$abcdefg$ is obviously divisible by 9. So just divide it by 9 to get suitable $y$.
For $x=3$, you have $y=37$ whose multiplication gives $111$.
I guess you can elaborate after this? 
A: Siong Thye Goh has a pretty cool solution but I have some kind of weird procedure that works too.
First, express $x=5^a\cdot2^b\cdot x'$. Since we can always multiply $x$ by $2^a\cdot 5^b$ to get $10^{a+b}x'$ and this is in $U$ precisely when $x'\in U$, we may assume that $\gcd(x,10)=1$ and just solve the problem.
Given $x\in \mathbb N^*$, are just looking for a positive integer $N=\sum_{i=0}^ka_i10^i$ with $a_i=0$ or $1$ such that $$\sum_{i=0}^ka_i10^i \equiv 0\pmod x. $$ (Once we find this, the $y$ in the problem will just be $N/x$.)
First write out the decimal expansion $x=\sum_{i=0}^d b_i10^i$ with $0\leq b_i\leq 9$. 
It is trivially true that $$\sum_{i=0}^d b_i10^i=x\equiv 0 \pmod x $$
Now we are going to find a number of the desired form (only has 0 and 1 in it's decimal expansion) congruent to this number.
Write each term $b_i 10^i$ as a sum of $b_i$ copies of $10^i$. Namely, $\overbrace{10^i+\cdots +10^i}^{b_i}$.
Since $(10,x)=1$, by Euler's theorem $(10^i)^{m\varphi(x)+1}=10^i$ for every non-negative integer $m$. Now we can replace all the $\sum_{i=0}^d b_i$ many powers of $10$ with distinct large powers of $10$. So when we take their sum, this will have only $0$s and $1$s in their decimal expansion because we are adding distinct powers of $10$.
The last two sentences also show that there are infinitely many "non boring" solutions, i.e. solutions that do not just differ from each other by multiplication by a power of $10$.
A: Here's a method which is similar to Siong Thye Goh's but for some $x$ yields a smaller result.
If $x=2^s5^t$, we may take $xy=10^{\max(s, t)}.$
Otherwise, let $b_i=10^i\bmod x$. By the pigeonhole principle there will be some $t$ and $s>t$ where $b_s=x-b_t$. One may then take $xy=10^s+10^t$.
For example if $x=7$, $(b_0,\dots)=(1, 3, 2, 6, 4, 5, 1,\dots)$ so we may take $t=0, s=3, xy=1001$.
