Find the units digit of floor(N), where:

10^{20000}
N = ---------
10^{100} + 3

*** For an extra challenge, solve this puzzle without using a computer program aided method.

I'm going to generalize this a little bit and make

N = 10^(2*h^2) / (10^h + 3)

For this problem h=100

Let M = 3^(2h)/(10^h + 3). M is less than 1. I will show that N-M is an integer, which means its last digit is the value KS asks us to find.

N-M = (10^(2*h^2) - 3^(2h)) / (10^h + 3)

Rearrange the exponents a little:

N-M = ((10^(2*h))^h - (3^2)^h)) / (10^h + 3)

This makes it easy to see that the numerator is of the form x^n-y^n. Then the numerator can be factored.

N-M = [10^(2*h) - (3^2)] * [(10^(2*h))^(h-1) + (10^(2*h))^(h-2)*(3^2) + ... + (10^(2*h))*(3^2)^(h-2) + (3^2)^(h-1)] / (10^h + 3)

The first term 10^(2*h) - (3^2) can be factored into (10^h - 3)*(10^h + 3). The second factor cancels the denominator in the fraction, leaving

N-M = (10^h - 3) * [(10^(2*h))^(h-1) + (10^(2*h))^(h-2)*(3^2) + ... + (10^(2*h))*(3^2)^(h-2) + (3^2)^(h-1)]

Since there are no more fractions, N-M must be an integer for positive integer h. The units digit is equal to (-3) * (-1)^(h-1) mod 10. This means when h is even the units digit is 3 and when h is odd the units digit is 7. h=100 is even, so __the digit KS asks for is 3__.