Is there any 3 digit no. with distinct digits which when reversed divides the original no. perfectly? Can there be such a number for any n digits?
4 years ago
The only numbers that work are 510, 540, 810.
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According to my calculations there are no such 3-digit numbers.
For this problem we need to find \(a\), \(b\) and \(c\), so that \(n*(100a+10b+c)=100*c+10b+a\).
To start off we look at the last and first digits:
\(n*c \equiv a\) (mod \(10\)) and \(n*a \lt 10\). We make a multiplication table and cross out all impossible combinations.
We are left with;
Here \(a=c\), so we don't have distinct digits.
Here \(a=2, c=6\) or \(a=4, c=7\).
Here \(a=2, c=4\) or \(a=1, c=7\)
Here \(a=2, c=3\) or \(a=2, c=8\)
Here \(a=1, c=9\).
Substituting these 8 possibilities in the equation from the second line gives us 8 equations in \(b\). None of these equations has a solution where \(0 \le b \le 9\).