HMMT November

Solution:

Let $\sqrt{n}$ simplify to $a_{n} \sqrt{b_{n}}$. Notice that both $a_{n}$ and $b_{n}$ are multiplicative. Thus, $\sum_{d \mid n} a_{d}$ and $\sum_{d \mid n} b_{d}$ are multiplicative.

We consider the sum $\sum_{d \mid p^{k}} a_{d}$ and $\sum_{d \mid p^{k}} b_{d}$. Notice that for $d=p^{l}$, $a_{d}=p^{\lfloor l / 2\rfloor}$ and $b_{d}=p^{2\{l / 2\}}$, so
$$\sum_{d \mid p^{k}} a_{d}=2\left(\frac{p^{(k+1) / 2}-1}{p-1}\right) \quad \text{ and } \quad \sum_{d \mid p^{k}} b_{d}=\frac{(p+1)(k+1)}{2}$$
for odd $k$, while
$$\sum_{d \mid p^{k}} a_{d}=\left(\frac{p^{(k+2) / 2}+p^{k / 2}-2}{p-1}\right) \quad \text{ and } \quad \sum_{d \mid p^{k}} b_{d}=\frac{(p+1) k}{2}+1$$
for even $k$.

Notice $8!=2^{7} \cdot 3^{2} \cdot 5 \cdot 7$, so
$$\sum_{d \mid 8!} a_{d}=\left(\frac{2(16-1)}{2-1}\right)\left(\frac{9+3-2}{3-1}\right)(1+1)(1+1)=30 \cdot 5 \cdot 2 \cdot 2=600$$
and
$$\sum_{d \mid 8!} b_{d}=\left(\frac{3 \cdot 8}{2}\right)\left(1+\frac{4 \cdot 2}{2}\right)(1+5)(1+7)=12 \cdot 5 \cdot 6 \cdot 8=2880$$
so the sum of $a_{d}+b_{d}$ would be $600+2880=3480$.