Binomial coefficients Congruence

Note that the expression is equal to ${102010 \choose 51005}$ . The base $101$ representations of these numbers are $A00$ and $500$ , where $A$ represents the single digit expression for $10$ . Thus, by Lucas' theorem,

${102010 \choose 51005} \equiv {10 \choose 5} \equiv 252 \equiv 50 \pmod {101}$ .

This problem involves a direct application of Lucas' Theorem. Essentially, the theorem gives a way of calculating large binomial coefficient modulo some prime p easily. It states that given two positive integers $m,n$ and a prime $p$ , the following congruence holds: $\binom{m}{n} \equiv \prod_{i = 0}^k \binom{m_i}{n_i} \pmod{p}$ , where $m = \sum_{i = 0}^k m_i p^i$ and $n = \sum_{i = 0}^k n_i p^i$ (this is the base $p$ representation of $m$ and $n$ ).

In the problem, we note that $\frac{102010!}{(51005!)^2} = \binom{102010}{51005}$ and that $102010 = 10(101)^2 + 0(101) + 0$ , $51005 = 5(101)^2 + 0(101) + 0$ . Applying Lucas' Theorem gives us $\binom{102010}{51005} \equiv \binom{10}{5} \binom{0}{0} \binom{0}{0} \equiv 252 \equiv 50 \pmod{101}$ . Thus, the answer is $\fbox{50}$ .

Note that $\frac{102010!}{(51005!)^2} = \binom{102010}{51005}$ and $102010= 10 \cdot 101^2$ $51005 = 5 \cdot 101^2$ Hence by Lucas' theorem we get $\binom{102010}{51005} \equiv \binom{10}{5} \equiv \boxed{50} \pmod{101}$