Certain $3$-digit numbers have the following characteristics:
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all three digits are different
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the number is divisible by $7$
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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?
Best Answer
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 $7$.
So the possible solutions are $168, 259, 861, 952$