Let $n$ be a three-digit number and let $m$ be the number obtained by reversing the order of the digits in $n$. Suppose that $m \neq n$ and that $n + m$ and $n − m$ are both divisible by $7$. Find all such pairs $n$ and $m$.
My idea was that you could say \begin{align}n=100a+10b+c\end{align} where $a$,$b$ and $c$ are digits $0-9$ and $a$ is not zero, therefore \begin{align} m &=100c+10b+a \\ n+m &=101a+20b+101c \\ n-m &=99a-99c \end{align}
If $7$ divides $n-m$ then $7$ divides $99a-99c=99(a-c)$, therefore $7$ must divide $a-c$, so the only possible combinations for $a$ and $c$ are $(9,2)$, $(8,1)$ and $(7,0)$ and $(2,9)$, $(1,8)$.
If $7$ divides $n+m=101a+20b+101c$ then $7$ divides $3a-b+3c=3(a+c)-b$.
If $(a,c)=(9,2)$ then $7$ has to divide $3(11)-b=33-b$ which gives $b$ possible values of $5$. Solution 1 and 2: $n=952$ or $n=259$.
If $(a,c)=(8,1)$ then $7$ has to divide $3(9)-b=27-b$ which gives $b$ possible values of $6$ . Solution 3 and 4: $n=861$ or $n=168$.
If $(a,c)=(7,0)$ then $7$ has to divide $3(7)-b=21-b$ which gives $b$ possible values of $0$ or $7$. Solution 5: $n=700$. Solution 6: $n=770$
Is there any others I have missed?