1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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\ $.

$\endgroup$
  • $\begingroup$ For example $3*33+2=101$ How can I be sure that this is the smallest number? And 14! doesn't disturb divisibility? $\endgroup$ – vmahth1 Oct 13 '19 at 18:40
  • $\begingroup$ 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!}\ $. $\endgroup$ – lonza leggiera Oct 14 '19 at 0:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.