The number of positive numbers less than 1000 and divisible by 5 (no digit being repeated) is

Here, the available digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

The numbers can be of one, two or three digits and in each of them unit’s place must have 0 or 5 as they must be divisible by 5.

The number of numbers of one digit = 1 (∵ 5 is the only number).

The number of numbers of two digits divisible by 5 = number of all the numbers of two digits divisible by 5 – number of numbers of two digits divisible by 5 and having 0 in ten’s place ${=}^2{\mathrm{P}}_1{\times }^9{\mathrm{P}}_1-1$,

(∵ unit’s place can be filled by either 0 or 5 in first category and only by 5 in the second category)

= 2 × 9 – 1 = 17.

The number of numbers of three digits divisible by 5 = number of all the numbers of three digits divisible by 5 = number of numbers of three digits divisible by 5 and having 0 in hundred’s place

${=}^2{\mathrm{P}}_1{\times }^9{\mathrm{P}}_2{-}^8{\mathrm{P}}_1\times 1=2\times 9\times 8-8=136$.

∴ Required number of numbers

= 1 + 17 + 136 = 154.