Articulated divisors

For a positive integer, its divisors come in pairs (e.g. $12=1\times 12=2\times 6=3\times 4;\: 16=1\times 16\times 2\times 8=4\times 4. )$ .

For a "strange" number, one pair of divisors need to be in the same number, for example, 16 (as mentioned above, it has a "repetitive" 4 as a divisor).

This requires the number to be squared.

$\because \textrm{INT}(\sqrt{2016})=44,$

$\therefore \textrm{Sum required}=\sum_{r=1}^{44}r^{2}=\frac{1}{6}\times 44\times (44+1)\times (2\times 44+1)= \boxed{29370}.$

Let's take $n=p_1^{q_1}p_2^{q_2}...$

First, we know that the number of divisors of $n$ is $d(n)=(q_1+1)(q_2+1)...$ . But if $d(n)$ in odd, that means that each of his factor is odd, so each of the $q_i$ is even and $n$ can be expressed as a perfect square.

We see that : $44^2 <2016 <45^2$ Hence, we have to calculate the sum of all the squares from $1$ to $44$ . Let's apply the formula : $\displaystyle\sum_{i=1}^{44}i^2=\frac {44\times (44+1)\times (2\times 44+1)}{6}=29370$ Hence the solution.

Reminded that the formula for the number of distinct positive divisors of the positive integer $n=\prod _{ l=1 }^{ j }{ p_l^{k_l} }$ is:

$\tau (n)=\prod _{ l=1 }^{ j }{ (k_l +1)}$

We know that $\tau (n)$ is odd, hence every single product term $k_{l\rightarrow j} +1$ is odd. In other words:

$k_{l\rightarrow j} +1 = 2y_{l \rightarrow j}-1$ for some integer $y$ that describes $k$ in correspondence

$\Leftrightarrow k_{l \rightarrow j} = 2(y_{l \rightarrow j}-1)$

Now we can plug the according value of $k_{l \rightarrow j}$ into the equation describing $n$ as the product of primes on certain powers:

$n=\prod _{ l=1 }^{ j }{ p_l^{2(y_l-1)} }$

$\Rightarrow n=(\prod _{ l=1 }^{ j }{ p_l^{y_l-1} })^2$

This suggests that $n$ is a perfect square. Therefore, the answer to this problem is to find the sum of all squares that are under or equal to $2016$ .

Knowing that $44^2 \le 2016 \le 45^2$ , we find the sum:

$\sum_{x=1}^{44}{x^2} = \frac{44(45)(89)}{6} = \boxed{29370}$