A Little Maths Puzzle

This puzzle appeared on PUZZLE A DAY and asks:

What is the only positive number that is equal to the average of its digits? Trailing zeros are not permitted. The trivial single digit answers don’t count as the question specifies ‘digits’. The clue provided is that the number consists of two digits. 

I made the mistake of thinking that the number is a two digit integer and reached an impasse. Let's represent the number with the digits \(x\) and \(y\). Then we have:$$ \begin{align} 10x + y &=\frac{x+y}{2} \\ 20x + 2y &=x+y \\ 19x + y &= 0 \end{align}$$Clearly this yields no solutions. However. Let's try the number \(x.y\) where the dot represents the decimal place. Then we have:$$ \begin{align} x + \frac{y}{10} &= \frac{x+y}{2} \\10x+y &= 5x+ 5y \\5x-4y &= 0 \end{align}$$This does yield a solution when \(x=4\) and \(y=5\). So the number must be 4.5 which indeed is the average of 4 and 5.

It's just a simple little problem but once you lock into that mindset that the number is an integer a solution seems impossible. I asked Gemini if there are positive numbers that satisfy for the geometric and harmonic means and the short answer is NO.