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0+1+2+3+4+5 = 15 so, if a 5-digit number, formed from these digits, is divisible by 3 it must exclude 0 or 3.

If it excludes 0, all 5*4*3*2*1 = 120 permutations are valid.

If it excludes 3, then 4*4*3*2*1 = 96 permutations are valid (the others start with a zero and so are 4-digit numbers).

That makes a total of 120+96 = 216 numbers.

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Q: How many 5 digits can be formed with digits 012345 divisible by 3 with digits not repeated?
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