Pro_GRMR wrote: ↑Tue Mar 30, 2021 12:46 am
The factor mean of a positive integer $n$ is equal to the geometric mean of its (positive) factors. If the number of positive integers $n$ whose factor mean is not greater than $34$ is $N$, then what is $N$?
Other variations contained $32$, $33$, or other values instead of $34$.
The process is similar using any of the values.
We know that the geometric mean of $k$ numbers $a_1, a_2, a_3, \dots, a_k$ is $\sqrt[^k]{a_1a_2 a_3\dots a_k}$.
Let's assume the number $n$ has $k$ factors $a_1, a_2, a_3, \dots, a_k$ where $a_{i-1} < a_i$.
If $n$ is not a perfect square, we can pair up its factors into groups and multiply $a_i$ and $a_{k+1-i}$ to get $n$ ($\overbrace{a_1a_k}^{n}\overbrace{a_2a_{k-1}}^{n}\dots\overbrace{a_\frac{k}{2}a_{\frac{k}{2}+1}}^{n}$). We will get $\frac{k}{2}$ such $n$.
Multiplying all of them we get $a_1a_2 a_3\dots a_k =\overbrace{a_1a_k}^{n}\overbrace{a_2a_{k-1}}^{n}\dots\overbrace{a_\frac{k}{2}a_{\frac{k}{2}+1}}^{n}= n^{\frac{k}{2}}$. We will get the same result even if $n$ is a perfect square.(Why?)
So, the factor mean of $n$ is $\sqrt[^k]{a_1a_2 a_3\dots a_k} = \sqrt[^k]{n^{\frac{k}{2}}} = \sqrt{n}$
According to the question:
$\sqrt{n} \leq 34$
$\Longrightarrow n \leq 34^2$
$\Longrightarrow n \leq 1156$
As $n$ is a positive number, there are a total of $\boxed{1156}$ possible values of $n$.