If you used a computer - then you're doing it wrong! We don't need that sort of overkill to solve this problem.
Write 10^99 as 10^6 * ((10^31)^3 + 27 - 27)
Then (10^99)/(10^31+3)
= (10^6 * ((10^31)^3 + 27 - 27))/(10^31+3)
Split this into two fractions:
10^6*((10^31)^3 + 27)/(10^31+3) - (27*10^6)/(10^31+3)
Using the sum of cubes factorization ((10^31)^3 + 27)/(10^31+3) can be simplified to (10^31)^2 - 3*10^31 + 9. This is an integer.
Then the first fraction reduces to the integer 10^6*((10^31)^2 - 3*10^31 + 9), which is clearly a multiple of 100.
The second fraction (27*10^6)/(10^31+3) is a very small positive fraction. Then the original expression (before flooring) is an integer minus a small positive fraction.
So the floor of the original expression is 10^6*((10^31)^2 - 3*10^31 + 9) - 1. This is one less than a multiple of 100, so the last two digits are 99.
Solution
