Consider a function $f\colon (0,1) \to \{0, 1\}$ defined as follows. For a real number $r \in (0,1)$ , $f(r) = 1$ if the second digit after the decimal point in $r$ is one of the four digits 2, 3, 6 and 7. Otherwise, $f(r)$ is equal to 0. The number of points in $(0,1)$ at which $f$ is discontinuous is __________. (answer in integer)

Any real number $r \in (0,1)$ can be represented in decimal form as $r = 0.d_1 d_2 d_3 \dots$, where $d_1 \in \{0, 1, 2, \dots, 9\}$ and $d_2 \in \{0, 1, 2, \dots, 9\}$.

The function is defined as:

For each fixed first digit $d_1 \in \{0, 1, \dots, 9\}$:

• $f(r) = 1$ when $r \in [0.d_1 2, 0.d_1 4)$ or $r \in [0.d_1 6, 0.d_1 8)$

• $f(r) = 0$ when $r \in [0.d_1 0, 0.d_1 2)$, $r \in [0.d_1 4, 0.d_1 6)$, or $r \in [0.d_1 8, 0.(d_1+1) 0)$

At the boundary points where $d_2$ changes between the subsets $\{2, 3\}$, $\{6, 7\}$ and the rest, the left-hand limit and right-hand limit of $f(r)$ differ (one limit is $0$ and the other is $1$), making $f$ discontinuous.

For each $d_1 \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, the boundary transition points are:

1. $0.d_1 2$

2. $0.d_1 4$

3. $0.d_1 6$

4. $0.d_1 8$

Each of these points lies strictly within the interval $(0, 1)$.

Since there are $10$ possible choices for $d_1$ and $4$ points of discontinuity for each choice, the total number of points in $(0,1)$ at which $f$ is discontinuous is: