A calculus problem by <> <>

$\begin{aligned} I(a) & = \int_0^{100} \{ax\} \ dx & \small \color{#3D99F6} \text{Let }u = ax \implies du = a \ dx \\ & = \int_0^{100a} \frac {\{u\}}a \ du \\ & = \frac 1a \sum_{n=1}^{100a} \int_0^1 u \ du \\ & = \frac 1a \sum_{n=1}^{100a} \frac 12 \\ & = 50 & \small \color{#3D99F6} \text{which is independent of }a \end{aligned}$

Therefore, we have:

$\begin{aligned} \sum_{a \in \mathbb S} \int_0^{100} \{ax\} \ dx & = 50 \color{#3D99F6} \sigma_0 (3600) & \small \color{#3D99F6} \text{where }\sigma_0 (\cdot) \text{ is the number of divisors function} \\ & = 50 \color{#3D99F6} ({\color{#D61F06}4}+1)({\color{#D61F06}2}+1)({\color{#D61F06}2}+1) & \small \color{#3D99F6} \text{since } 3600 = 2^{\color{#D61F06}4} \times 3^{\color{#D61F06}2} \times 5^{\color{#D61F06}2} \\ & = \boxed{2250} \end{aligned}$

The function $\{x\}$ has a period of $1$

$\Longrightarrow \{ax\}$ has a period of $\frac{1}{a}$

Here $a=3$

Hence to evaluate the area between $0$ and $100$ we have to sum the areas of $100a$ small triangles. Since all of them have the same hight of $1$ , their area combined is

$100\cdot 1\cdot \dfrac{1}{2}=50$

Thus the required sum is

$M=\displaystyle\sum_{a\in \mathbb{S}} \int_0^{100} \{ax\}dx=\displaystyle\sum_{a\in \mathbb{S}} 50=50\cdot |\mathbb{S}|$

Where $|\mathbb{S}|$ is the cardinality of $\mathbb{S}$ . Since $\mathbb{S}$ is the set of divisors of $3600=2^4\cdot 3^2\cdot 5^2$ , its cardinality is $(4+1)(2+1)(2+1)=45$

Hence

$M=45\cdot 50=\boxed{2250}$