The numbers are lying between 100 and 1000 then each number is of three digits.

The possible combination of 3 digits numbers are

1, 2, 3; 1, 2, 4; 1, 2, 5; 1, 3, 4; 1, 3, 5; 1, 4, 5; 2, 3, 4; 2, 3, 5; 2, 4, 5; and 3, 4, 5.

The possible combination of numbers which are divisible by 3 are 1, 2, 3; 3, 4, 5; 1, 3, 5 and 2, 3, 4.
(If sum of digits of a number is divisible by 3 then the number is divisible by 3)

$$ \therefore $$ Total number of numbers = 4 × 3! = 24

The possible combination of numbers divisible by 5 are 1, 2, 5; 2, 3, 5; 3, 4, 5; 1, 3, 5; 1, 4, 5 and 2, 4, 5.
(If the last digit of a number is 0 or 5 then the number is divisible by 5)

$$ \therefore $$ Total number of numbers = 6 × 2! = 12

The possible combination of number divisible by both 3 and 5 are 1, 3, 5 and 3, 4, 5.

$$ \therefore $$ Total number of numbers = 2 $$ \times $$ 2! = 4

$$ \therefore $$ Total required number = 24 + 12 - 4 = 32