There are: \(9-1+1=9\left(numbers\right)\) from 100 to 199.

There are: \(9-2+1=8\left(numbers\right)\) from 200 to 299.

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There are: \(9-9+1=1\left(number\right)\) from 900 to 999.

So there are total is: \(9+8+7+6+5+4+3+2+1=45\left(numbers\right)\)  satisfy the question.

there are:9-1+1=9(numbers)

there are :9-2 +1=8(numbers)

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there are:9-9+1=1(number)

the number of number have 3 digits can disivible by 3 is:

9+8+7+6+5+4+3+2+1=45 (numbers)

answer:45 numbers