TITLE: Finding n and k that satisfies ${n \choose k} = 517 $
QUESTION [12 upvotes]: How do I find all the integers n and k that satisfies ${n \choose k} = 517 $ ? How do I approach this question?

REPLY [4 votes]: ${n \choose 0}$ is always 1, there is no solution
${n \choose 1} = n$, which gives the solution ${517 \choose 1}$
${n \choose 2} = \frac{n(n-1)}{2}$, which leads to the $n^2-n-1034=0$ quadratic equation, which doesn't have an integer solution.
${n \choose 3} = \frac{n(n-1)(n-2)}{6}$. But ${15 \choose 3}=455$, and ${16 \choose 3}=560$, so there is no solution again.
Similarly, hogy ${n \choose 4}$, we got ${12 \choose 4}=495$ and ${13 \choose 4}=715$.
For ${n \choose 5}$, ${11 \choose 5}=462$ and ${12 \choose 5}=792$. Here is no solution, too.
For ${n \choose 6}$, ${11 \choose 6}=462$ (just as above), but ${12 \choose 6}=924$.
From this point on, if we found a solution ${n \choose k}$, we had to also find it before. (Why?)

Thus the only solutions are ${517 \choose 1}$ and ${517 \choose 516}$.