$\begin{array}{1 1} 100 \\ 60 \\ 40 \\ 24 \end{array}$

For a 3 digit number to be divisible by 3, sum of its digits

should be divisible by 3.

With 0 as one of the digits the remaining 2 digits can be (1,2),(1,5),(2,4),(4,5)

After choosing the three numbers they can be arranged in 3! ways.

But with 0 as one of the digits, 0 cannot come in hundreds place for 3 digit numbers..

$\therefore$ No. of numbers with 0 as one of its digits = $4\times (3!-2!)=16$

With 1 as one of the digits the remaining 2 digits can be $ (2,3),(3,5)$

$\therefore$ No. of numbers with 1 as one of its digits $=2\times 3!=12$

With 2 as one of the digits the remaining 2 digits are (3,4)

$\therefore $No. of numbers with 2 as one of its digits =$1\times 3!=6$

With 3 as one of the digits the remaining 2 digits are (4,5)

$\therefore$No. of numbers with 3 as one of its digits $= 1\times 3!=6$

Total numbers $= 16+12+6+6=40$

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