How many such three digit numbers are possible? Certain $3$-digit numbers have the following characteristics:


*

*all three digits are different

*the number is divisible by $7$

*the number on reversing the digits is also divisible by $7$
How many such numbers are there? I have tried using a brute force approach and found $168$ and $259$. Is there a better way to solve these questions?
 A: We can say that the number has distinct digits $a,b,c$ and then we can write it as $$10^2a+10b+c$$
Now we have \begin{align}\left(10^2 a+10 b+c\right)\mod 7&\equiv 0\tag{1}\\
\left(10^2 c+10 b+a\right)\mod 7&\equiv 0\tag{2}\end{align}
Now we consider $(2)-(1)$:
\begin{align}\left(10^2 c+10 b+a\right)-\left(10^2 a+10 b+c\right)&=10^2(c-a)+10(b-b)+(a-c)\\
&=10^2(c-a)-(c-a)\\
&=99(c-a)\end{align}
Can you continue from here?
A: Let 
\begin{align}
N_1 &= 10^{2}a+10b+c \\
N_2 &= 10^{2}c + 10b +a
\end{align}
Whence
\begin{align}
N_1-N_2 &= 100a-100c+10b-10b+c-a \\
&= 100(a-c)+(a-c) \\
&= 99(a-c)
\end{align}
Thus the difference between your two numbers should also be divisible by $9$. 
A: If the number is $10^2a+10b+c$ as digits then considering $(10^2a+10b+c) - (10^2c+10b+a)$ leads to realising $99(a-c)$ must be divisible by $7$ and cannot be zero and, since $99$ is coprime to $7$.


*

*$1b8$ but $108 \equiv 3 \pmod 7$ requiring $10b \equiv 4 \pmod 7$ making the only possible value of $b=6$, which is neither $1$ nor $8$ $\checkmark$  

*$2b9$ but $209 \equiv 6 \pmod 7$ requiring $10b \equiv 1 \pmod 7$ making the only possible value of $b=5$, which is neither $2$ nor $9$ $\checkmark$  

*$7b0$ but $700 \equiv 0 \pmod 7$ requiring $10b \equiv 0 \pmod 7$ making the only possible values of $b=0$ or $7$, which are either $7$ or $0$ $\times$ 

*$8b1$ but $801 \equiv 3 \pmod 7$ requiring $10b \equiv 4 \pmod 7$ making the only possible value of $b=6$, which is neither $8$ nor $1$ $\checkmark$   

*$9b2$ but $902 \equiv 6 \pmod 7$ requiring $10b \equiv 1 \pmod 7$ making the only possible value of $b=5$, which is neither $9$ nor $2$ $\checkmark$  


So the possible solutions are $168, 259, 861, 952$
