It is not all that hard, if we remember the frequently useful $2^{10}=1024$, which is $2.4$ percent more than $1000$.
The small power of $2$ that is closest to beginning with $7$ is $2^6$, which is $64$. We need to increase $64$ by a bit under $10$ percent, times a power of $10$. But since we are concerned only about the first digit, we don't really even see the powers of $10$, intuitively we just want to push from $6.4$ to $7$.
It is easy to see that pushing up twice by $2.4$ percent is not enough to push us up to beginning with $7$. That doing it three times is not enough is less obvious, but a bit of fooling around, even without a calculator, does the job: although the compounding helps, $2.4$ percent three times does not give $10$ percent. It is clear that increasing $64$ by $2.4$ percent four times will get us to the $7$, and leaves us quite far from beginning with $8$. So the required power is $6+(10)(4)$.
The starting digit that takes longest to reach is $9$. This is trickier to do by hand, for $2^{53}$ barely gets us there. It is the first legitimate candidate, but calculations have to be more accurate than for the digit $7$, to make sure that we don't need to go up to $63$.
Claim 1:
The divisibility rule for a number '$a$' to be divided by '$n$' is as follows. Express the number '$a$' in base '$n+1$'. Let '$s$' denote the sum of digits of '$a$' expressed in base '$n+1$'. Now $n|a \iff n|s$. More generally, $a \equiv s \pmod{n}$.
Example:
Before setting to prove this, we will see an example of this. Say we want to check if $13|611$. Express $611$ in base $14$.
$$611 = 3 \times 14^2 + 1 \times 14^1 + 9 \times 14^0 = (319)_{14}$$
where $(319)_{14}$ denotes that the decimal number $611$ expressed in base $14$. The sum of the digits $s = 3 + 1 + 9 = 13$. Clearly, $13|13$. Hence, $13|611$, which is indeed true since $611 = 13 \times 47$.
Proof:
The proof for this claim writes itself out. Let $a = (a_ma_{m-1} \ldots a_0)_{n+1}$, where $a_i$ are the digits of '$a$' in the base '$n+1$'.
$$a = a_m \times (n+1)^m + a_{m-1} \times (n+1)^{m-1} + \cdots + a_0$$
Now, note that
\begin{align}
n+1 & \equiv 1 \pmod n\\
(n+1)^k & \equiv 1 \pmod n \\
a_k \times (n+1)^k & \equiv a_k \pmod n
\end{align}
\begin{align}
a & = a_m \times (n+1)^m + a_{m-1} \times (n+1)^{m-1} + \cdots + a_0 \\
& \equiv (a_m + a_{m-1} \cdots + a_0) \pmod n\\
a & \equiv s \pmod n
\end{align}
Hence proved.
Claim 2:
The divisibility rule for a number '$a$' to be divided by '$n$' is as follows. Express the number '$a$' in base '$n-1$'. Let '$s$' denote the alternating sum of digits of '$a$' expressed in base '$n-1$' i.e. if $a = (a_ma_{m-1} \ldots a_0)_{n-1}$, $s = a_0 - a_1 + a_2 - \cdots + (-1)^{m-1}a_{m-1} + (-1)^m a_m$. Now $n|a$ if and only $n|s$. More generally, $a \equiv s \pmod{n}$.
Example:
Before setting to prove this, we will see an example of this. Say we want to check if $13|611$. Express $611$ in base $12$.
$$611 = 4 \times 12^2 + 2 \times 12^1 + B \times 12^0 = (42B)_{12}$$
where $(42B)_{14}$ denotes that the decimal number $611$ expressed in base $12$, $A$ stands for the tenth digit and $B$ stands for the eleventh digit.
The alternating sum of the digits $s = B_{12} - 2 + 4 = 13$. Clearly, $13|13$. Hence, $13|611$, which is indeed true since $611 = 13 \times 47$.
Proof:
The proof for this claim writes itself out just like the one above. Let $a = (a_ma_{m-1} \ldots a_0)_{n+1}$, where $a_i$ are the digits of '$a$' in the base '$n-1$'.
$$a = a_m \times (n-1)^m + a_{m-1} \times (n-1)^{m-1} + \cdots + a_0$$
Now, note that
\begin{align}
n-1 & \equiv (-1) \pmod n\\
(n-1)^k & \equiv (-1)^k \pmod n \\
a_k \times (n-1)^k & \equiv (-1)^k a_k \pmod n
\end{align}
\begin{align}
a & = a_m \times (n-1)^m + a_{m-1} \times (n-1)^{m-1} + \cdots + a_0 \\
& \equiv ((-1)^m a_m + (-1)^{m-1} a_{m-1} \cdots + a_0) \pmod n\\
a & \equiv s \pmod n
\end{align}
Hence proved.
Pros and Cons:
The one obvious advantage of the above divisibility rules is that it is a generalized divisibility rule that can be applied for any '$n$'.
However, the major disadvantage in these divisibility rules is that if a number is given in decimal system we need to first express the number in a different base. Expressing it in base $n-1$ or $n+1$ may turn out to be more expensive. (We might as well try direct division by $n$ instead of this procedure!).
However, if the number given is already expressed in base $n+1$ or $n-1$, then checking for divisibility becomes a trivial issue.
Best Answer
$x=1\cdot 10^n + t$, with $t < 10^n$
$y=10t+1 \implies y=3x$
$10t+1 = 3 \cdot 10^n + 3t \implies 7t=3 \cdot 10^n-1$
So the question is: what is the smallest $n$ such that $7$ divides $3 \cdot 10^n-1$ ?
Another way to express this is: how long is the period in $3/7$ when expressed in decimal?
This will give you the smallest $x$.