Factorials are going to drive you crazy!

We use the congruence ${ap \choose bp} \equiv {a \choose b} \mod p$ .

Setting $p = 101$ and $a = 2b = 1010$ gives ${102010 \choose 51005} \equiv {1010 \choose 505} \mod 101$ .

Applying the congruence again gives ${102010 \choose 51005} \equiv {10 \choose 5} \equiv \boxed{50} \mod 101$ .

By Lucas' theorem: Since (102010)!/(51005!)^2 = 102010 C 51005 = (102010 51005), we can use Lucas' theorem for modulo 101. Since 102010 = 10 x 101^2 and 51005 = 5 x 101^2, 102010 C 51005 is congruent to 10 C 5 = 252 and is congruent to 50 (mod 101).

We use Lucas' theorem: Since (102010)!/(51005!)^2 = 102010 C 51005 = (102010 51005), we can use Lucas' theorem for modulo 101. Since 102010 = 10 x 101^2 and 51005 = 5 x 101^2, 102010 C 51005 is congruent to 10 C 5 = 252 and is congruent to 50 (mod 101).