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;

\(n=1\)

Here \(a=c\), so we don't have distinct digits.

\(n=2\)

Here \(a=2, c=6\) or \(a=4, c=7\).

\(n=3\)

Here \(a=2, c=4\) or \(a=1, c=7\)

\(n=4\)

Here \(a=2, c=3\) or \(a=2, c=8\)

\(n=7\)

Here \(a=1,c=3\)

\(n=9\)

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 \(0 \le b \le 9\).

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TopNewestThe 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;

\(n=1\)

Here \(a=c\), so we don't have distinct digits.

\(n=2\)

Here \(a=2, c=6\) or \(a=4, c=7\).

\(n=3\)

Here \(a=2, c=4\) or \(a=1, c=7\)

\(n=4\)

Here \(a=2, c=3\) or \(a=2, c=8\)

\(n=7\)

Here \(a=1,c=3\)

\(n=9\)

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 \(0 \le b \le 9\).

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