This equation was drawn to my attention with the instruction "solve for $n$."


Most of this is easy. Take the umpth root, multiply by the denominator, stuff like that. My only issue is finding a method to remove the factorial operator whilst keeping the other side of the equal sign satisfied.

Is there a quick method of doing this? Is there something easier than trying every possible combination and adjusting the solution to fit?

Any help would be appreciated.



As soon as $n!$ exceeds $6^n$, which is $n=12,$ the rest of the terms won't matter because they are too small. Actually the $6^n$ is big enough to matter up to $n=14$. That isn't many to try. Then $\frac n6$ is non-integer unless $n=0,6,12$ so you can only try those three. Only $0$ works. Done.

  • $\begingroup$ Well, you want $n!$ to be not just bigger than $6^n$, but substantially bigger. $\endgroup$ – Robert Israel Jun 6 '17 at 4:25
  • $\begingroup$ Technically it doesn't have to be integer, I've heard something about the factorial being expanded to non-integer values $\endgroup$ – Alex Li Jun 6 '17 at 4:30
  • $\begingroup$ @AlexLi: there is the gamma function which is a real function of a real argument (or a complex function of a complex argument) but when it is written as factorial it is fair to assume it is natural numbers. Then all the terms in the equation are natural with the possible exception of $\frac n6$ $\endgroup$ – Ross Millikan Jun 6 '17 at 4:32
  • $\begingroup$ @Ross Millikan fair enough. I was just thinking that the problem wanted an actual solution ("solve for n" implies, to me, that there is a solution). Also, I just noticed, -1 could potentially give an integer if the 6^n was being added. Negative whole factorials should be okay. $\endgroup$ – Alex Li Jun 6 '17 at 4:37
  • $\begingroup$ Actually negative whole factorials (based on the gamma function) are infinite because of $(n-1)!=\frac {n!}n$ going through $n=0$ so they do not apply. I just noticed that $n=0$ is a solution, so there is one, and updated. $\endgroup$ – Ross Millikan Jun 6 '17 at 4:40

Hint : $n! >> 6^n > n^6 > n$ for large enough $n$


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