remainder question for big number Every 2 digit positive integer is repeated 5 times and strung together (ex. 1818181818). Find the remainder when all of these integers are added together and divided by 505

Not sure best way to do it -
If you repeat the two digit number $n=10a+b$ (ie digits $a,b$) five times you get $ababababab$, which is $n\times 101010101$. It is easy to check that $101010101=1\bmod505$. So the sum of all these 10 digit numbers mod 505 is the same as the sum of all two digit numbers mod 505.

$10+\dots+99$ is an AP with sum $45\times109$ (as usual pair the numbers $10+99=109,11+98=109$ etc and multiply by half the number of terms). 

It is easy to check that $45\times109=360\bmod505$.
A mechanistic way to write your initial sum is
$$
S = \sum_{n=10}^{99} \sum_{k=0}^{4} n10^k
$$
Then your question reduces to
$$
S \equiv x \mod{505}
$$

Can you solve this relationship?