What is the Smallest Integer $N$ Where Reversing the Digits Makes $3N$? What is the smallest positive integer N such that the integer formed by reversing the digits of N is triple N? (Does such an integer even exist? If not, then for what multiplier for $N$ will such an integer exist?)
Here are my thoughts so far: (where $n$ is the number of digits of $N$)
    $$N=\sum\limits_{i=0}^{n}{d_i 10^i}$$
    $$3N=\sum\limits_{i=0}^{n}{d_i 10^{n-i}}$$
therefore $$3\sum\limits_{i=0}^{n}{d_i 10^i}=\sum\limits_{i=0}^{n}{d_i 10^{n-i}}$$
so $$\sum\limits_{i=0}^{n}{d_i(10^{n-i}-3\times10^i)}=0$$
$$d_0(10^n-3)+d_n(1-3\times10^n)+\sum\limits_{i=1}^{n-1}{d_i(10^{n-i}-3\times10^i)}=0$$
$$d_0(10^n-3)+d_n(1-3\times10^n)+10\sum\limits_{i=1}^{n-1}{d_i(10^{n-i-1}-3\times10^{i-1})}=0$$
Then using congruence relations,
$$d_0(10^n-3)+d_n(1-3\times10^n)+10\sum\limits_{i=1}^{n-1}{d_i(10^{n-i-1}-3\times10^{i-1})}\equiv 0\pmod{10}$$
$$d_0(10^n-3)+d_n(1-3\times10^n)\equiv  0\pmod{10}$$
However, this doesn't seem like the right way to go; even if I can get many congruence relations I would still have to brute-force many different N values to confirm the congruence relations. So how can I solve the puzzle without brute-forcing it?

There is no such $N$ (besides $0$). 
Now, say $N$'s first digit is $a$ and its last digit is $b$. Since $N$ and $3N$ have the same number of digits, $a$ can only be $1, 2$, or $3$. If $a=3$, then $b=9$ is the only choice, but this is impossible since then $3N$ would end in $7$, not $3$. $a=1$ also doesn't work, since then $b$ would have to be $3, 4,$ or $5$, and none of those triple to $1 \mod 10$. Similarly, $a=2$ doesn't work, since then $b$ must be $6, 7,$ or $8$, none of which triple to a number ending in $2$.   
As an extension, you can show by the same reasoning that it still doesn't work if you change $3$ to any positive integer greater than $1$, except possibly $4$ (where $N$ must start with $2$ and end with $8$) and $9$ (where $N$ must start with $1$ and end with $9$). It obviously won't work if the multiplier is greater than $9$ since then the number of digits will change.
