Summer's over

The only such values of $n$ are $9, 18$ and $24$ , giving a sum of $51$ . Now for a quick proof of why these are the only such values of $n$ .

Let $n = \prod_{i=1}^m{(p_{i})^{a_{i}}}$ be the prime factorization of $n$ .

Now by induction we can show that $2^{k} \ge (k + 1)$ for all natural numbers $k$ . Since $2$ is the least prime we can then say that $p^{k} \ge (k + 1)$ for all primes $p$ and natural $k$ .

Now suppose $n$ has $7$ as a prime factor. We can show by induction that $7^{k} \gt 3(k + 1)$ for all natural $k$ . Then

$n = 7^{a_{j}} * \prod_{i \ne j}^m{(p_{i})^{a_{i}}} \gt 3(a_{j} + 1) * \prod_{i \ne j}^m{(a_{i} + 1)} = 3*d(n)$ .

This implies that if $n$ has $7$ as a prime factor then $n \ne 3*d(n)$ . This is in fact the case for all primes $\ge 7$ , and so the prime factorization of $n$ is of the form $2^{a} * 3^{b} * 5^{c}$ for some non-negative integers $a,b,c$ .

Now we can prove by induction that $5^{k} \gt 3(k + 1)$ for $k \ge 2$ , so $c = 0$ or $c = 1$ .

We can also prove by induction that $3^{k} \gt 3(k + 1)$ for $k \ge 3$ , and so $b = 1$ or $b = 2$ . (Note that $n$ must be divisible by $3$ so $b \ne 0$ .)

Yet another induction proof gives us that $2^{k} \gt 3(k + 1)$ for $k \ge 4$ , and so $a$ has possible values of $0, 1, 2$ and $3$ .

(Sorry I'm leaving out the proofs by induction, but this solution is long enough as it is.)

Now if $c = 1$ , then we would have $2^{a} * 3^{b} * 5 = 3*(a + 1)*(b + 1)*2$ . This would imply that $5$ divides either $(a + 1)$ or $(b + 1)$ . But given the bounds we've established for $a$ and $b$ this is impossible, so we must have $c = 0$ . Thus $n = 2^{a} * 3^{b}$ .

Now if $b = 1$ then

$2^{a} * 3^{b} = 3(a + 1)(b + 1) \Longrightarrow 3*2^{a} = 6(a + 1) \Longrightarrow 2^{a-1} = a + 1 \Longrightarrow a = 3$ .

This gives us $n = 24$ as a solution.

If $b = 2$ then we would have

$2^{a} * 9 = 9(a + 1) \Longrightarrow 2^{a} = a + 1 \Longrightarrow a = 0,1$ .

This gives us solutions $n = 9$ and $n = 18$ .

Thus the only possible values for $n$ are $9, 18$ and $24$ , giving us a sum of $\boxed{51}$ .

Solution in 2 rows. The number of divisor of $n$ between $\frac{n}{4}$ and $n$ is less or equal than $4$ . So $\frac {n}{4}+4>\frac {n}{3}$ . It implies that $n<48$ . We know that $3\mid n$ , so we only need check the cases.

Another solution is to notice that $d(n)$ can never be greater than $\lfloor\sqrt{n}\rfloor \times 2$ because for every pair of factors of $n$ one must be above and one must be below $\sqrt{n}$ (both can be equal to $sqrt{n}$ but that complicates the solution).

$d(n)$ = $\sqrt{n} \times 2$ at numbers such as $n = 2,6,12$ .

For $d(n) = \frac{n}{3}$ to be true $\lfloor\sqrt{n}\rfloor \times 2 \geq \frac{n}{3}$ must be true.

If $\lfloor\sqrt{n}\rfloor \times 2 \geq \frac{n}{3}$ is false for a perfect square (or equality exists) no value of $n$ higher than the perfect square exists. Therefore, we need to find the highest perfect square for which this is true. Because it is a perfect square, we have: $\lfloor\sqrt{n}\rfloor = \sqrt{n}$ . So we have:

$\sqrt{n} \times 2 \geq \frac{n}{3}$

$\sqrt{n} \times 6 \geq n$

The highest value for which this is true is at $n = 36$ . As equality exists, it is not true for $n > 36$ . Now we only must check the values of $n$ below 37 which are divisible by 3.

We find $n=9,18$ and $24$ are the only solutions the answer is $\boxed{51}$ .

We can extend this to the general case: $d(n) \times k = n$ , where $k$ is a fixed integer.

The maximum number of solutions for $n$ can be found by finding the highest square for which:

$\sqrt{n} \times 2k \geq n$ is true, which is at $n = (2k)^{2}$ .

$n$ must not be greater than $(2k)^{2}$ and must also be divisible by $k$ so the maximum value of solutions to $d(n)$ is $4k$ . I believe it is possible to prove that the number of solutions cannot be greater than a value lower than $4k$ but cannot prove it myself.

I welcome any corrections/additions.

Let, $n=3^{k} \prod p_{i}^{q_{i}}$ where $p_{i }$ is any particular prime number excluding 3. Now, $d(n)=(k+1) \prod (q_{i}+1)$ The numbers we are trying to find must satisfy $d(n)=\frac{n}{3}$ or, $(k+1) \prod (q_{i }+1)=3^{k-1} \prod p_{i }^{q_{i }}$ or, $\frac{k+1}{3^{k-1}}=\frac{\prod p_{i }^{q_{i }}}{\prod (q_{i }+1)}$ ...(1)

Now, $\frac{\prod p_{i }^{q_{i }}}{\prod (q_{i }+1)} \geq 1$ where equality occurs when all $q_{i }=0$ or when $\prod p_{i }^{q_{i }}=2^1$ . Hence, the following must be true. $\frac{k+1}{3^{k-1}} \geq 1$ This happens for $k \leq 2$ . As we can see $k=2$ makes LHS of equation (1) equal to 1. How many ways can we make RHS equal to 1? In two ways, to be precise. One way is to make all $q_{i}=0$ and another way is to set $\prod p_{i }^{q_{i }}=2^1$ . These two cases respond to two numbers $3^2=9$ and $3^2 2^1=18$ . For $k=1$ the LHS becomes equal to 2. To get 2 from RHS we need to have 2 as a prime in $\prod p_{i }^{q_{i }}$ . A little exploration shows that $\prod p_{i }^{q_{i }}=2^3$ can make the RHS exactly equal to 2 and this corresponds to the number $3^1 2^3=24$ . Therefore, the answer is $9+18+24=\boxed{51}$ .