B3. The lobster of a positive integer is the number that results when you write its digits in reverse order. Thus, the lobster of 8267 is 7628 and the lobster of 15620 is 2651 (since leading zeros always drop off).

What is the smallest positive integer $n$ such that $n$ minus the lobster of $n$ equals 12345678?
B3. 20406080 It is clear that $n$ must be at least 12345678. To find the smallest possible $n$, we start by trying eight-digit numbers. Write $n$ as abcdefgh, the number with the digits a, ..., h. The reverse of $n$ is then $k=\overline{\text { hgfedcba }}$ and $n-k=12345678$.

$$
\begin{aligned}
& \text { abcdefgh } \\
& \frac{\text { hgfedcba }}{12345678}-
\end{aligned}
$$

To get the smallest possible $n$, we try to make the first digit as small as possible. We try $\mathrm{a}=1$. If we then look at the first digit of $n-k$, we see that $\mathrm{h}=0$ must hold, but then the last digit of $n-k$ is 9, which is incorrect. Therefore, we see that $\mathrm{a} \geqslant 2$ must hold.

We take $\mathrm{a}=2$ and for the second digit, we try the smallest possible digit: $\mathrm{b}=0$. By looking at the last two digits of $n-k$, we see that $\mathrm{g}=8$ and $\mathrm{h}=0$ must hold. So far, we have

Since the first and last two digits are already correct, as long as nothing needs to be borrowed from the second digit, we now look for $c$, $d$, $e$, and $f$ such that $\overline{c d e f}-\overline{f e d c}=3456$.

The smallest $\mathrm{c}$ we can choose now is $\mathrm{c}=3$. If we do that, we find $f=0$ by looking at the third digit of $n-k$. The sixth digit of $n-k$ then becomes a 7, which is incorrect. Therefore, we see that $\mathrm{c} \geqslant 4$ must hold.

We take $\mathrm{c}=4$ and the smallest possible $\mathrm{d}$, namely $\mathrm{d}=0$. By looking at the fifth and sixth digits of $n-k$, we find that $\mathrm{e}=6$ and $\mathrm{f}=0$ must hold. We now find the solution $n=20406080$, which indeed satisfies $20406080-8060402=12345678$.