, 2015

Solution:

Answer: 127

Since the value of each digit is restricted to $\{1,2, \ldots, 7\}$, there is exactly one digit representative of each residue class modulo $7$.

Note that $7 \mid \overline{a b c}$ if and only if $100 a + 10 b + c \equiv 0 \pmod{7}$ or equivalently $2a + 3b + c \equiv 0$. So we want the number of triples of residues $(a, b, c)$ such that at least one of $2a + 3b + c \equiv 0$, $2b + 3c + a \equiv 0$, $2c + 3a + b \equiv 0$ holds.

Let the solution sets of these three equations be $S_1, S_2, S_3$ respectively, so by PIE and cyclic symmetry we want to find $3|S_1| - 3|S_1 \cap S_2| + |S_1 \cap S_2 \cap S_3|$.

Clearly $|S_1| = 7^2$, since for each of $a$ and $b$ there is a unique $c$ that satisfies the equation.

For $S_1 \cap S_2$, we may eliminate $a$ to get the system $0 \equiv 2(2b + 3c) - (3b + c) = b + 5c$ (and $a \equiv -2b - 3c$), which has $7$ solutions (one for each choice of $c$).

For $S_1 \cap S_2 \cap S_3 \subseteq S_1 \cap S_2$, we have from the previous paragraph that $b \equiv -5c$ and $a \equiv 10c - 3c \equiv 0$. By cyclic symmetry, $b, c \equiv 0$ as well, so there's exactly $1$ solution in this case.

Thus the answer is $3 \cdot 7^2 - 3 \cdot 7 + 1 = 127$.