Find the smallest number for which the number divides completely into 101 
Find the smallest number $n,(n>4)$,
  $A=\binom{3n-1}{11}+\binom{3n-1}{12}+\binom{3n}{13}+\binom{3n+1}{14}$
  for which the number divides completely into $101$.

My solution:
$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}\\\binom{3n-1}{11}+\binom{3n-1}{12}=\binom{3n}{12}\\\binom{3n}{12}+\binom{3n}{13}=\binom{3n+1}{13}\\\binom{3n+1}{13}+\binom{3n+1}{14}=\binom{3n+2}{14}\\\binom{3n+2}{14}=\frac{(3n+2)!}{14!(3n+2-14)!}=\frac{(3n+2)!}{14!(3n-12)!}$ 
and at the moment I don't know how to do the task efficiently

Hint: You're almost there.  Since $101$ is a prime number, it will divide $\ \frac{(3n+2)!}{14!(3n-12)!}\ $ evenly if and only if it divides one of the numbers $\ 3n+2,3n+1, 3n, \dots,3n-11\ $.
