Least count of Vernier

Is least count not always 1 main scale divsion-1vernier scale divsion or is it always least reading possible on main scale/ no of vernier divsions. question: 1 main scale reading be 1 mm Lets say 5 main scale reading coincide with 8 vernier scale division then what is least count? 1-5/8= 3/8 mm or 1/8 mm.

if for above vernier scale, the Main scale reading is 35, and 4th divison of vernier coincides with main scale. (zero error=0). then why is this wrong: Main scale reading + Least count* Vernier scale reading 35+ 3/8 *4 =36.5

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 $$V3$$.