Crazy divisors

If $k \neq n$ is a factor of $n^2$ , then so is $n^2/k$ . This means that, other than $n$ , the factors of $n^2$ come in pairs.

If $k < n$ then $\frac{n^2}{k} > n$ and vice-versa, so we can conclude that half of the factors (different than $n$ ) are greater than $n$ and half are less than $n$ .

Since $n^2 = 2^{62} \times 3^{38}$ , it has $(62+1)(38+1) - 1 = 2456$ factors other than $n$ , of which $\frac{2456}{2} = 1228$ are less than $n$ .

Now we just have to subtract the ones that are also factors of $n$ . Since $n = 2^{31} \times 3^{19}$ , it has $(31+1)(19+1) - 1 = 639$ factors other than $n$ . Subtracting the two we have $1228 - 639 = 589$ .

i) let's start how many dividers $n^2$ have. : $\tau(n^2)=(62+1)*(38+1)=2457$ ii) than let's think how many divder's is $<n$ : $(\tau(n^2)-1)/2=(2457-1)/2=1228$ (cause except $n$ , we can pair the divider which is $<n$ and $>n$ $a*b=n^2$ iii) also, let's think how many $n$ 's divder is $<n$ : $\tau(n)-1=(31+1)*(19+1)-1=639$ (except $n$ ) Answer is ii)-iii) $1228-639=589$ $■$

The divisors of $n^2$ less than $n$ but do not divide $n$ are divisors with power of $2$ , $p > 31$ or power of $3$ , $q > 19$ and $2^p3^q < n$ .

From $2^p3^q < n \Longrightarrow p \log{2} + q \log{3} < \log{n}$ , we find that the largest $p$ is $\left \lfloor \frac{\log{n}}{\log{2}}\right \rfloor = 61$ and that of $q$ is $\left \lfloor \frac{\log{n}}{\log{3}}\right \rfloor = 38$ . We note that when $q = 38, 37, 36$ the largest $p_{q}= 0,2,4$ , therefore the qualified divisors with $3^{38}$ , $3^{37}$ and $3^{36}$ as factors are $1(0+1)$ , $3(2+1)$ and $5(4+1)$ respectively. There are corresponding $q_p$ for each $p$ . Therefore, the total number of these divisors are given by:

$\begin{aligned} N & = \sum_{q=20}^{38} {(p_q+1)} + \sum_{p=32}^{61} {(q_p+1)} \\ & = \sum_{q=20}^{38} {\left( \left \lfloor \frac{\log{n}-q \log{3}}{\log{2}} \right \rfloor +1 \right)} + \sum_{p=32}^{61} {\left( \left \lfloor \frac{\log{n}-p \log{2}}{\log{3}} \right \rfloor +1 \right)} \\ & = 297 + 292 = \boxed{589} \end{aligned}$