Three digit number divisibility by 7

Let $A = 100a + 10b + c$ and $B = 100c + 10b + a$. $B - A = 99(c - a)$. Since $B > A$, $c > a$. $B - A$ is divisible by 7, and 99 is not divisible by 7, so $(c - a)$ must be divisible by 7. Since $a$ and $c$ are single-digit non-zero numbers, the possible values for $(c, a)$ are $(9, 2)$ or $(8, 1)$.

• If $a = 1, c = 8$, then $A$ ranges from $108$ to $198$.
• If $a = 2, c = 9$, then $A$ ranges from $209$ to $299$. Hence $A$ ranges from $108$ to $299$. This range lies completely within $106 < A < 305$.