# Is there a number $n>2$ that has $n$ factors, including itself

I was wondering if there is any number $$n>2$$ such that it also has $$n$$ factors, including itself. Clearly $$1$$ and $$2$$ both meet the requirements, which is why I set $$n>2$$. I came across highly composite numbers on Wikipedia, but I didn't find exactly what I was looking for. If there is no such $$n>2$$, is there a proof showing why it is not possible?

• n factors implies that the number is at least 2^n Commented Dec 30, 2019 at 2:04
• n=2 has 2 factors but $2<2^2$ Commented Dec 30, 2019 at 2:06

No. To have $$n$$ factors every number from $$1$$ to $$n$$ would have to divide $$n$$. For $$n \gt 2, \ n-1$$ does not divide $$n$$.
You might be interested in Euler's totient function, where $$\phi(n)$$ is the number of numbers $$k$$ less than $$n$$ which are coprime to $$n$$, that have $$\gcd(k,n)=1$$. All of these except $$1$$ are nonfactors of $$n$$ but there can be numbers that are neither coprime nor factors.
• Yes, that seems so obvious I do not know why I did not think of that. This is what I've worked out from what you have said: $\frac{n}{n-1} \text{(an integer if n-1|n)}=\frac{n-1+1}{n-1}=1+\frac{1}{n-1} \text{(an integer iff n=2)}$. Commented Dec 30, 2019 at 1:28