A vernier caliper has a least count of 0.05 mm and the value of one...

Given, Least Count (LC) $= 0.05\text{ mm}$ and $1\text{ MSD} = 0.5\text{ mm}$.

The least count is given by the difference between one main scale division and one vernier scale division:
$\text{LC} = 1\text{ MSD} - 1\text{ VSD}$
$0.05\text{ mm} = 0.5\text{ mm} - 1\text{ VSD}$
$1\text{ VSD} = 0.45\text{ mm}$

The total length of the vernier scale, which has $20$ divisions, is:
$\text{Total length} = 20 \times 1\text{ VSD} = 20 \times 0.45\text{ mm} = 9.0\text{ mm}$

Let $M$ be the number of main scale divisions that coincide with the entire vernier scale.
Length of $M$ main scale divisions $= M \times 1\text{ MSD} = M \times 0.5\text{ mm}$

Equating the lengths:
$M \times 0.5 = 9.0$
$M = \dfrac{9.0}{0.5} = 18$

Thus, $18$ main scale divisions coincide with $20$ vernier scale divisions.

Answer: $18$