The last two digits of  $$3^{2019}$$ are ?​

Correct option is D

To find the last two digits required we have to calculate $$3^{2019} \equiv p \ mod100$$.

From Euler's theorem: if a(prime) and 100 are co prime then the satisfy:

​$$a^{\phi(n)} \equiv 1mod \ n$$ , where $$\phi(n)$$ is Euler's phi function​

Using this, we get.

​$$3^{\phi(100)}=1mod100\\[10pt]\phi(100)=40\\[10pt]\implies 3^{40}\equiv 1 mod100\\[10pt]\implies (3^{40})^{50}\equiv1^{50} mod100\\[10pt]\implies 3^{2000}\equiv 1mod 100\\[10pt]\implies 3^{2019}\equiv 1.3^{19}mod100\\[10pt]\implies 3^{2019}\equiv 13^{19}mod100\\[10pt]\textbf{Now compute :}3^{19} mod100\\[10pt]3^4=81, 3^{8}=6561\equiv 61 mod 100\\[10pt]\implies 3^{16}\equiv 61^{2}mod100\\[10pt]\implies3^{16}\equiv 3721mod 100\\[10pt]\implies 3^{16}\equiv 21mod 100\\[10pt]\implies 3^{16}.3^{3}\equiv 21.3^3mod 100\\[10pt]\implies 3^{19}\equiv 567mod 100\\[10pt]\implies 3^{19}\equiv 67mod 100\\[10pt]\textbf{Hence, last two digits will be 67}\\[10pt]\implies \textbf{Option D is correct.}$$​​