What are the last two digits of $3^{100}11^{50}+7^{518}$? What are the last two digits of $3^{100}11^{50}+7^{518}$?

I know that to find the last two digits implies that you are essentially finding the remainder when divided by $100$. 

Can I simplify $3^{100}$ to $3$ mod $1$ by dividing both the exponent and $100$ (which are the same in this case) by $100$?

Also, how would I simplify $7^{518}$ if that were the case?

Any help is appreciated; thanks!
As you suggested, we want some $0\leq n \leq 99$ such that
$$ 
n \equiv 3^{100} \cdot 11^{50} + 7^{518} \pmod{100}.
$$
As noted by lulu in the comments, we first observe that $3^{100} = 9^{50}$. Hence, we have
\begin{align}
n &amp;\equiv 3^{100} \cdot 11^{50} + 7^{518} \pmod{100} \\ &amp;= 9^{50} \cdot 11^{50} + 7^{518} \\ &amp;= 99^{50} + 7^{518}
\end{align}

Note that $99^2 \equiv 1 \pmod{100}$ and $7^4 \equiv 1 \pmod{100}$, so
\begin{align}
n &amp;\equiv 99^{50} + 7^{518} \pmod{100}\\ &amp;= (99^2)^{25} + (7^4)^{129} \cdot 7^2 \\ &amp;\equiv 1^{25} + 1^{129} \cdot 7^2  \pmod{100} \\&amp; = 1 + 49 =50. \end{align}

Therefore, the last two digits of $3^{100} \cdot 11^{50} + 7^{518}$ are $5$ and $0$.
$$ 3^{100}11^{50} = 9^{50}11^{50} = 99^{50}$$

$$ 99^{50} (mod 100) = (-1)^{50} (mod 100) = ((-1)^2)^{25} = 1^{25} = 1$$

$$ 7^4 (mod 100) = 2401 (mod 100) = 1 $$

$$ 7^{518} (mod 100)= 7^2(7^{4})^{129} (mod 100) = 7^2 = 49 $$

$$ 1 + 49 = 50 $$

Then, the result is: $$50$$