Last Two digits!

Pascal’s triangle has a property where if all the digits in the $n$ th row are read, it gives the $n$ th power of 11. ${2018 \choose 2018}=1$ , giving the last digit. ${2018 \choose 2017}={2018 \choose 1}=2018$ , so the last digits are ...20181. This leaves 81 as the last two digits.

Let $N=11^{2018}$ . We need to find $N \bmod 100$ .

$\begin{aligned} N & \equiv 11^{2018} \text{ (mod 100)} \\ & \equiv (10+1)^{2018} \text{ (mod 100)} \\ & \equiv \left(10^{2018} + 2018 \times 10^{2017} + \cdots + \frac {2018\times 2017}2 \times10^2 + 2018\times 10 + 1\right) \text{ (mod 100)} \\ & \equiv \boxed{81} \text{ (mod 100)} \end{aligned}$

Notice that:

$11^2 \mod 100 =21$ ,

$11^3 \mod 100 =31$ ,

$11^4\mod 100=41\dots \text { etc }$

$\implies 11^{2018} = 11^{2010+8} \mod 100 = (1)(81)= 81 \mod 100 \implies \text { the last two digits are 81. }$

$11 ^ { 2018 } = 1 ^ { 8 }$ when we look for the last digit, so last digit must be $1$ .

Then we only need to look for the last-second digits.

It is $8$ .

So, it is $\boxed { 81 }$ .