maths question 2

You should know one simple thing to shortlist the possible number of choices.

Every 3-digit number, which is divisible by 3, must have the sum of 3 digits also divisible by 3.

You can use this fact and choose only the odd digits to minimize the number of possibilities and use simple counting to find the answer.

可被3整除的有:
111,117,135,153,159,171,177,195,
315,333,339,357,351,375,393,399,
513,519,531,537,555,573,579,591,597,
711,717,735,753,759,771,777,795,
915,933,939,951,957,915,993.
So th answer is C.39