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Note that $p|n$ .So let $n=p^{r}p_{1}^{r_{1}}....p_{k}^{r_{k}}$ where $p_{i}$ form a strict increasing sequence of primes.

Divide the problem into 2 cases

i. $k=0$ .

Hence, $p^{r-1}=r+1$ .

The only solutions are $p=2,r=3$ and $p=3,r=2$ .

Hence, $n=8$ or $9$

ii. $k≥1$ .Now consider the following two sub-cases-

a. $p_{1}≥3$

Then $p_{i}^{r_{i}}>r_{i}+1$

Now, $p^{r-1}p_{1}^{r_{1}}....p_{k}^{r_{k}}=(r+1)(r_{1}+1)....(r_{k}+1)$ .

So, $p^{r-1}<r+1$ .This occurs only when $r=1$ or $(p,r)=(2,2)$ .

Let $r=1$ .Then $2|p_{1}^{r_{1}}....p_{k}^{r_{k}}$ contradiction!

So, $(p,r)=(2,2)$ is the only choice.

Hence, $2.p_{1}^{r_{1}}...p_{k}^{r_{k}}=3(r_{1}+1)...(r_{k}+1)$

Hence, $p_{1}=3$ . Hence $2.3^{r_{1}}...p_{k}^{r_{k}}=3(r_{1}+1)...(r_{k}+1)$

We know that, $2.3^{r_{1}}≥3(r_{1}+1)$ .Equality occurs iff $r_{1}=1$ .

Also, $p_{i}^{r_{i}}≥r_{i}+1$ for $i≥2$

So, we require $k=1$ and $r_{1}=1$ .Hence $n=2^{2}.3^{1}=12$

b.Let $p_{1}=2$ .Now again consider two sub-cases.

i. $r=1$ .

Then, $2^{r_{1}}≤ 2(r_{1}+1)$ , Therefore, $r_{1}=1,2 or 3$

However, $r_{1}=1$ or $3$ implies $2|\frac{n}{pp_{1}}$ i.e., $2|p_{2}$ . But $p_{2}<p_{1}=2$ .

So, then n cannot have any prime factor, i.e., $k=1$ .

But then, $r_{1}=1$ => $p^{0}=2$ => $1=2$

$r_{1}=3$ => $8.p^{0}=4(r+1)$ => $r=1$

So, $r_{1}=1$ leads to contradiction and $r_{1}=3$ leads to circularity which is to say that any prime $p≠2$ satisfies it, which implies $n=8p$ for some prime $p≠2$

If $r_{1}=2$ , then $p_{2}=3$ . Then similarly as above, $2.3^{r_{2}}≤3(r_{2}+1)$ which is satisfied by only $r_{2}=1$ and hence $n$ has no other prime factor, i.e., $k=2$ .

Hence, $n=12p$ for any prime $p>3$ .

ii. $r>1$

Then, $p^{r-1}≥r+1$ and $p_{i}^{r_{i}}>r_{i}+1$ for all $i>1$

Therefore, $2^{r_{1}}≤r_{1}+1$ which has $r_{1}=1$ as the only solution.But then again the inequality above becomes an equality and so n cannot have any other prime factor.

This makes $k=1$ and $p^{r-1}=r+1$ which occurs only when $p=3$ and $r=2$ .

So, $n=18$ is another solution.

In summary,

1) If $p=2$ , hen $n=8$ or $n=12$

2)If $p=3$ , then $n=9$ , $n=18$ or $n=24$

3)If $p>3$ , then $n=8p$ or $n=12p$

Now count the number of all such $n$ .The answer will be $103$ .