Find the integers that double if their first and last digits are swapped.
In other words find the digits: $a$, $b$ and $n_1$ ... $n_m$ in such a way that:
[$a$ $n_1$ $n_2$ ... $n_m$ $b$] $\times$ $2$ = [$b$ $n_1$ $n_2$ ... $n_m$ $a$]
For example, using the rule above: $25$ turns into $52$ but unfortunately $52 \gt 25 \times 2$. The number $102$ transforms into $201$ but again $102 \times 2 \gt 201$.
UPDATE
Possible solution:
Case 1: the number ends in $1$ and in consequence its double ends in $2$
$\ldots 1 \times 2$
$\ldots2$
In conclusion the pair must be:
$2 \ldots 1 \times 2$
$1 \ldots 2$ (impossible)
Case 2: the number ends in $2$
$4 \ldots 2 \times 2$
$2 \ldots 4$ (impossible)
Case 3:
$6 \ldots 3 \times 2$
$3 \ldots 6$ (impossible)
Case 4:
$8 \ldots 4\times 2$
$4 \ldots 8$ (impossible)
Case 5:
$0 \ldots 5 \times 2$ (the first digit can not be zero)
$5 \ldots 0$
Case 6:
$2 \ldots 6 \times 2$
$6 \ldots 2$ (impossible)
Case 7:
$4 \ldots 7 \times 2$
$7 \ldots 4$ (impossible)
Case 8:
$6 \ldots 8 \times 2$
$8 \ldots 6$ (impossible)
Case 9: the number ends in $9$
$8 \ldots 9 \times 2$
$9 \ldots 8$ (impossible)
Case 10: the number ends in $0$
$0 \ldots 0 \times 2$ (the number can not start with $0$)
$0 \ldots 0$
I'll give an algebraic proof. Your answer is good, though.
Given a number $l$, greater than $10$ since this is obviously not true for single digit numbers, write it as $l = 10^nx + 10y + z$, where $x$ and $z$ are single digit numbers i.e. $10^n \leq l < 10^{n+1}$. Essentially, we are isolating the first and last digits separately (for example, $166465 = 1 \times 10^5 + 6646 \times 10 + 5$)
Now, from the given condition, $2l = 10y + 10^n z + x$, since we have only swapped the last two digits. Of course, by ordinary multiplication, this is also equal to $2 \times 10^n x + 20y + 2z$, by multiplying the expression for $l$ by $2$.
Consequently, $10y + 10^n z + x = 2\times 10^n x + 20y + 2z$. A few transpositions give $10y = (10^n - 2) z - (2 \times 10^n-1)x$.
The last digit of the left hand side is $0$. Therefore, so is the last digit of the number on the right hand side. Note that $n \geq 1$, since we know $l$ has two digits. This gives the last digit of the right hand side as the last digit of $x - 2z$.
Consequently, $x - 2z$ must have last digit zero. That is, either $x = 2z$, or $x = 2z - 10$ (note that $-20 < x - 2z < 10$).
But $y \geq 0$, so $z \geq \frac{(2 \times 10^n - 1) x}{10^n - 2} \geq 2x \geq x$. So $x = 2z - 10$ is the only possibility : the other forces $x \geq z$.
But if $x = 2z - 10$ then $10 y = 10^n z - 2z - 2(10^n)(2z - 10) + 2z - 10 = - 3 \times 10 ^n z + 2 \times 10^{n+1} - 10$, and dividing by $10$ gives $y = 2 \times 10^n - 3 \times 10^{n-1}z - 1 = 10^{n-1}(20- 3z) - 1$.
This forces $z \leq 6$ from $y \geq 0$. Also, note that $2z - 10 \geq 0$ so $z \geq 5$.
Note that $z = 6$ is ruled out since this gives $x = 1$ but $x$ is the last digit of an even number, hence is even.
But if $z = 5 $ then $x = 0$, but $x \neq 0$ by assumption that $10^n \leq l$.
Thus, no such $l$ exists.