reduce this to $12^{12}$ by euler's totient theorem. Bashing thru some numbers, we obtain $12^{12} \equiv 56 \pmod{100}$

The answer is 56. This is a solution that does not require the Euler Totient Theorem. Since we are finding the last two digits of $2012^{2012},$ we can just take the modulo 100 of this number to obtain $2012^{2012}\pmod{100}\equiv12^{2012}\pmod{100}$ Now, observe that $12^1\pmod{100}\equiv12\pmod{100}$ $12^2\pmod{100}\equiv44\pmod{100}$ $12^3\pmod{100}\equiv28\pmod{100}$ $12^4\pmod{100}\equiv36\pmod{100}$ $12^5\pmod{100}\equiv32\pmod{100}$ $12^6\pmod{100}\equiv84\pmod{100}$ $12^7\pmod{100}\equiv8\pmod{100}$ Therefore, we can simplify $12^{2012}\pmod100$ as $(8^{287}\pmod{100})(12^3\pmod{100})\equiv(8^{287}\pmod{100})(28\pmod{100})$ We can continue computing in this way to find that $8^1\pmod{100}\equiv8\pmod{100}$ $8^2\pmod{100}\equiv4\pmod{100}$ $8^3\pmod{100}\equiv2\pmod{100}$ $8^4\pmod{100}\equiv6\pmod{100}$ $8^5\pmod{100}\equiv8\pmod{100}$ This is a recurrence. Hence, we can now find that $(8^{287}\pmod{100})(28\pmod{100})\equiv(8^3\pmod{100})(28\pmod{100})\equiv(2\pmod{100})(28\pmod{199})\equiv(56\pmod{100}).$ We can conclude that our final answer is 56 Hope this helps! Please tell me if there are any errors in my solution.