The answer is no. Suppose the answer would be affirmative. Obviuously the three digits cannot be all of the same parity, because we would not be able to cover all the residues modulo 2 Thus there exists at least one even and one odd digit. Suppose WLOG that we have one odd digit and two even digits (the other case can be treated in the same way, replacing the word even with the word odd everywhere). Thus the 8 odd residues modulo 16 must be covered by odd numbers, that is the last digit is the odd one. But there are 9 ways to choose the first two digits, and out of those at least 4 must give odd residue when divided by 8, because the number $\overline{xy}\cdot 10$ ( $\overline{xyz}=\overline{xy}\cdot 10 +z\ \Rightarrow\ \overline{xy}\cdot 10$ ) must pass through all the even residues modulo 16. The last 4 numbers which give odd residue when divided by 8 must be formed having the last digit y, and odd number, but there are only 3 such numbers, which leads to a contradiction. Thus the problem is solved. I am not sure, but I belive this was my solution in the contest.

why there are at least 4 odd residues modulo 8 and why xy.10 must pass through all even residues modulo 16? (any help is appreciated)

guesswho1729 wrote: why xy.10 must pass through all even residues modulo 16? According to the solution above, when there is exactly one odd digit, and the last digit is odd, 8 odd numbers $\overline{xyz}=\overline{xy}\cdot 10 +z$ gives the distinct odd residues 1, 3, 5, 7, 9, 11, 13, 15. So $\overline{xy}\cdot 10$ gives the distinct even residues 0, 2, 4, 6, 8, 10, 12, 14. guesswho1729 wrote: why there are at least 4 odd residues modulo 8 From the 8 cases of distinct even residues mentioned above, $\overline{xy}\cdot 10=16a+0\Rightarrow \overline{xy}\cdot 5=8a+0\Rightarrow \overline{xy}$ is even. $\overline{xy}\cdot 10=16a+2\Rightarrow \overline{xy}\cdot 5=8a+1\Rightarrow \overline{xy}$ is odd. $\overline{xy}\cdot 10=16a+4\Rightarrow \overline{xy}\cdot 5=8a+2\Rightarrow \overline{xy}$ is even. $\overline{xy}\cdot 10=16a+6\Rightarrow \overline{xy}\cdot 5=8a+3\Rightarrow \overline{xy}$ is odd. $\overline{xy}\cdot 10=16a+8\Rightarrow \overline{xy}\cdot 5=8a+4\Rightarrow \overline{xy}$ is even. $\overline{xy}\cdot 10=16a+10\Rightarrow \overline{xy}\cdot 5=8a+5\Rightarrow \overline{xy}$ is odd. $\overline{xy}\cdot 10=16a+12\Rightarrow \overline{xy}\cdot 5=8a+6\Rightarrow \overline{xy}$ is even. $\overline{xy}\cdot 10=16a+14\Rightarrow \overline{xy}\cdot 5=8a+7\Rightarrow \overline{xy}$ is odd. So there are exactly 4 odd numbers for $\overline{xy}$. That happens when $y$ is an odd digit. But there are only 3 choices for $x$, which leads to a contradiction.