15. Find the number of three-digit numbers which are multiples of 3 and are formed by the digits $0,1,2,3,4,5,6,7$ without repetition.
15. Answer: 106
Solution. Note that an integer is a multiple of 3 if and only if the sum of its digits is a multiple of 3. Also note that the sum of three integers $a, b, c$ is a multiple of 3 if and only if either (i) $a, b, c$ all have the same remainder when divided by 3 , or (ii) $a, b, c$ have the distinct remainders when divided by 3 . Observe that the remainders of $0,1,2,3,4,5,6,7$ when divided by 3 are $0,1,2,0,1,2,0,1$ respectively.
For case (i), the only possible selections such that all the three numbers have the same remainder when divided by 3 are $\{0,3,6\}$ and $\{1,4,7\}$. With $\{0,3,6\}$, we have 4 possible numbers (note that a number does not begin with 0 ), and with $\{1,4,7\}$, there are 6 possible choices.
For case (ii), if the choice of the numbers does not include 0 , then there are $2 \times 3 \times 2 \times 3!=72$; if 0 is included, then there are $3 \times 2 \times 4$ choices.
Hence the total number of possible three-digit numbers is $72+24+10=106$.