remainder

41

using fermat's theorem a^x≡1(mod n) x= number of integers < n which are prime to n
gcd(a,n)=1
now x for 100 will be 40 and gcd (3,100)=1 ..so we can apply this theorem

3^2012≡[3^40.3^40.3^40...to 50 terms ]. 3^12
≡1.3^12(mod100)
≡41 ..if my calculations are correct

\hspace{-16}$Here $\bf{3^{2012}=9^{1006}=\left(1-10\right)^{1006}}$\\\\\\ $\bf{\left(1-10\right)^{1006}=\binom{1006}{0}-\binom{1006}{1}.(10)+\binom{1006}{2}.(10)^2+.............}$\\\\\\ $\bf{\left(1-10\right)^{1006}=1-1006\times 10+\mathbb{M}(100)}$\\\\\\ Where $\bf{\mathbb{M}(100)=}$ Multiple of $\bf{100}$\\\\\\ So $\mathbf{\frac{\left(1-10\right)^{1006}}{100}=\frac{-10059+\mathbb{M}(100)}{100}=\frac{-59}{100}=\frac{\left(100-59\right)-100}{100}=41}$\\\\\\ So Remainder is $\bf{41}$\\\\\\ Or We can Use Congruencies Modulo........

Brilliant Nkhlshd.. Using Fermat Theorem

remember dis fact....in powers of 3 and 7 the last two digits repeat itself after a cycle of 20(which we formally call as cyclicity....) since 3²⁰ can be written as 9¹⁰ or (10-1)¹⁰ and in the binomial expansion of this all terms have 2 zeroes except the last term which is (-1)¹⁰....so last two digits of 3²⁰ is 01....(same is the case for 7 also)
now 3²⁰≡01(mod 100)
using this we can find the last two digits of any power of 3 or 7!!!