# 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\$$.
• For example $3*33+2=101$ How can I be sure that this is the smallest number? And 14! doesn't disturb divisibility? – vmahth1 Oct 13 '19 at 18:40
• If $101$ divides one of the numbers $\ 3n+2,3n+1,3n, \dots,$$3n-11\$, then $\ 3n+2\$ cannot be smaller than $101$. Therefore, $\ \frac{(3n+2)!}{14!(3n-12)!}\$ cannot be smaller than $\ \frac{101\cdot100\cdot\,\dots\,\cdot88}{14!}\$. The only prime factors of $\ (3n+2)(3n+1)\dots(3n-11)\$ which can be cancelled out by the $\ 14!\$ are $\ 2,3,5,7,11\$ and $\ 13\$, so if $101$ is a factor of $\ (3n+2)(3n+1)\dots(3n-11)\$ it will still be a factor of $\ \frac{(3n+2)(3n+1)\dots(3n-11)}{14!}\$. – lonza leggiera Oct 14 '19 at 0:17