prove that (72!/36!36!) -1 divisible by 73

or (72C36)-1 is divisible by 73

For 0

You can also prove the identity

C(n,k) = sum_j=0..k (-1)j+kC(n+1,j)

Edit:

Or if you understand modular arithmetic, with C(73,k)=0 (mod 73) you can expand by the binomial recurrence C(n,k)=C(n-1,k)+C(n-1,k-1) to get

C(72,k) + C(72,k-1) = 0 (mod 73)

C(72,k) = -C(72,k-1) = 0 (mod 73)

and then iterating

C(72,k) = (-1)kC(72,0) = (-1)k (mod 73)