The problem is as follows:

"By a mistake a number was doubled instead of raising it to the power of two. The result was a two digit number which was the reverse of what it should have been. What was the original number?"

I realized rapidly that the solution must be a number between 5 and 9. Any less than 5 and doubling does not result in a two digit number, any more then 9 and squaring does not result in a two digit number.

Clearly the answer is 9 since $2·9 = 18$ and $9^2 = 81$. [Simple, well below mypay grade.]

But is there an algebra solution to this problem?

I got as far as this simple system...

$\displaystyle 2n = 10a + b \Longrightarrow n = 5a + \frac{b}{2}$

$\displaystyle n^2 = 10b + a$

Substituting $n$ from the top equation into the bottom one leads to nothing useful.

$\displaystyle (5a + \frac{b}{2})^2 = 10b + a \Longrightarrow 100a^2 + 20ab + b^2 = 40b + 4a$

Am I missing some clever way of solving the problem using algebra?

Thanks.