I can calculate the number with computer and read the digits, or I can say PowerMod[2, 100, 100] in Mathematica, or I can even determine the digits myself, pencil-and-paper, by considering congruences modulo 4 and modulo 25 separately.
But if I determine the last digit first -- it's 6 -- and then set this congruent equation for the penultimate digit $y$:
$\frac{2^{100}-6}{10} \equiv y\ (mod\ 10)$
Which simplifies to $2^{99} - 3 \equiv 5\ y\ (mod\ 10)$, or $8 \equiv 5\ y + 3\ (mod\ 10)$, or $y \equiv 1\ (mod\ 2)$. This says this digit is odd. But how can I determine its exact value from here?
I wonder why sequential modulo 10 congruences don't yield a specific answer.
As another method, we have :$2^{20}\equiv1(\bmod25)\implies2^{100}\equiv1(\bmod25)$ from the euler's theorem. Now, it is trivially true that $2^{100}\equiv0(\bmod4)$. Now, you can use the chinese remainder theorem to find the remainder modulo $100$ which is your required answer. For further specifics, you can refer this answer.
The answer is $76$ [$2^{100}\equiv4(25)(1)+1(19)(4)(\bmod100)$ because of the fact that $25(1)\equiv1(\bmod4) $ and $4(19)\equiv1(\bmod25)$]