How many five digit numbers divisible by $3$ can be formed using the digits $0,1,2,3,4,7$ and $8$ if each digit has to be used at most once.

We've to make $5$ digit numbers using given 7 digits excluding 2 digits every time.

Now,$0+1+2+3+4+7+8=25$.

As 25\  mod\  3 = 1, duplets to be excluded should have their sum mod\  3 = 1.

It's not much difficult to find such duplets. Without much trouble,

we get $:(0,1),(0,4),(0,7),(1,3),(2,8),(3,4),(3,7)$.

Thus, there are such 7 duplets & hence 7 such groups of 5 digits.

Now, we find that out of 7 groups, 4 groups have 0 in them which isn't allowed to take 5th place ( otherwise a 4-digit number will be formed.).

Total numbers formed by these 4 groups $=4(5!-4!)=384$.

Other 3 groups have no zero so total numbers formed by them$=3(5!)=360$

Total possibel numbers $384+360=744$