If the last 3 digits of $2012^m$ and $2012^n$ are identical, find the smallest possible value of $m+n$.

You have $12^m\equiv 12^n\pmod{10^3}$ hence $12^m\equiv 12^n\pmod{2^3}$ and $12^m\equiv 12^n\pmod{5^3}$.

For the latter, we have $12^{m-n}\equiv 1\pmod{5^3}$, hence $m\equiv n\pmod{100}$ because the multiplicative order of $12$ modulo $5^3$ is $100$, as computed here.

On the other hand, $12^n(12^{m-n}-1)\equiv 0\pmod{2^3}$ from which $12^n\equiv 0\pmod{2^3}$ which holds for $n\geq 2$.

Thus $n=2$ and $m=102$ is the smallest solution with sum $m+n=104$.

Let find periodicity of last digit: $$ 1\underline{2}, 14\underline{4}, 172\underline{8}, 2073\underline{6}, 24883\underline{2}, ... $$ Since $12^1 \equiv 12^5 \equiv 2 \bmod(10)$, we conclude that $12^k\equiv 12^{k+4} (\bmod 10)$.

Now find periodicity of last $2$ digits, using this $4$-periodicity of last digit: since $12^4 \equiv 36 (\bmod 100)$,

$12^1 =12 (\bmod 100)$,
$12^5 \equiv 12 \cdot 12^4 \equiv 12\cdot 36 \equiv 32 (\bmod 100)$,
$12^9 \equiv 12^5 \cdot 12^4 \equiv 32\cdot 36 \equiv 52 (\bmod 100)$,
$12^{13} \equiv 12^9 \cdot 12^4 \equiv 52\cdot 36 \equiv 72 (\bmod 100)$,
$12^{17} \equiv 12^{13}\cdot 12^4 \equiv 72\cdot 36 \equiv 92 (\bmod 100)$,
$12^{21} \equiv 92\cdot 36 \equiv 12 (\bmod 100)$;

so, periodicity of last $2$ digits is $20$.

Finally, find periodicity of last $3$ digits: since $12^{20} \equiv 176 (\bmod 1000)$, we have:

$12^2 \equiv 144 (\bmod 1000)$;
$12^{22} \equiv 144 \cdot 176 \equiv 344 (\bmod 1000)$;
$12^{42} \equiv 344 \cdot 176 \equiv 544 (\bmod 1000)$;
$12^{62} \equiv 544 \cdot 176 \equiv 744 (\bmod 1000)$;
$12^{82} \equiv 744 \cdot 176 \equiv 944 (\bmod 1000)$;
$12^{102} \equiv 944 \cdot 176 \equiv 144 (\bmod 1000)$;

So, $$12^2 = 144,$$ $$12^{102} \equiv 144 (\bmod 1000).$$

$$(m+n = 104)$$