You have not designed a "conventional" vernier scale.
Let the distance between two consecutive main scale divisions (pitch) be $D$ and the distance between two vernier scale divisions be $V$.
The connection between $D$ and $V$ is $(n-1)D = nV$ with $\frac Dn$ and $D-V$ being called the least count.
So you can have $5\, (=n-1)$ main scale divisions equal to $6\, (=n)$ vernier divisions but "you need to work harder" if you have $5\, (=n-3)$ main scale divisions equal to $8\, (=n)$ vernier divisions.
Here is a picture of your main scale $D$ and vernier scale $V3$ with $5$ main scale divisions equal to 8 vernier scale divisions.
The vernier scale $V3$ is moved to the position where the $1$ on the vernier scale first coincides with a main scale graduation $(36).
The vernier has moved $\frac 38$ of a main scale division and so the reading is $35 \frac 38$ with the least count being $\frac 3 8$.
Now keep moving the vernier passing coincidence between a main scale mark and $2$ and $3$ on the vernier scale until coincidence with $4$ on the vernier scale is reached.
The vernier has moved along $4 \times \frac 3 8 = 1\frac 12$ and the reading is $35 + 1\frac 12 = 36.5$
Now the vernier scale can be relabeled as shown for $V1$ in red.
Starting at a reading of $35$ move the vernier scale until the first coincidence of any vernier scale division with the main scale.
The vernier has moved $\frac 18$ and the reading is $35\frac 18$ with the least count being $\frac 18$.
Keep moving the vernier scale passing coincidence with red $2$ and red $3$ to have coincidence between the main scale and red $4$ which happens to be the same label as that on vernier scale $V3$.
The vernier scale has now moved $4 \times \frac 18 = \frac 12$ and the reading is $35\frac 12$.
Note that the scale reading on $V1$ is equal to modulo $8$ of three times a scale reading on $V3$.