4. Each of $A$ and $B$ is a four-digit palindromic integer, $C$ is a three-digit palindromic integer, and $A-B=C$.
What are the possible values of $C$ ?
[A palindromic integer reads the same 'forwards' and 'backwards'.]
SOLUTION
Let the integers be $A=$ ' $a d d a$ ', $B=$ 'beeb' and $C=$ ' $c f c$ '. Then we may rewrite $A-B=C$ in the form of the addition sum shown.

Note that none of $a, b$ and $c$ can be zero, because $A, B$ and $C$ are integers with 4,4
beeb and 3 digits.
$\frac{+c f c}{a d d a}$
Considering the 'ones' column, $a \neq b$ since $c \neq 0$. Considering the 'thousands' column, since $a \neq b$, there is a 'carry' from the 'hundreds' column. It follows that $a=b+1$, so that, from the 'ones' column, $c=1$. Once we know that $c=1$, the only way there can be a 'carry' from the 'hundreds' to the 'thousands' column is to have $e=9$ or $e=8$.

There are two cases.
$e=9$
In this case, we have the addition shown alongside.
Considering the 'hundreds' column, $d$ is either 0 or 1 , depending on whether there is not, or is, a 'carry' from the 'tens' column. But when $d=0$ there is such a 'carry' $(f=1)$, which leads to a contradiction.
Hence $d=1$ and thus $f=2$, as shown alongside.
$\begin{array}{r}b 99 b \\ +1 f 1 \\ \hline a d d a\end{array}$
$e=8$
We now have the addition shown alongside.
b99b
$\frac{+121}{a 11 a}$
In this case, we similarly have $d=0$ and $f=2$, as shown alongside.
$b 88 b$
$\frac{+1 f 1}{a d d a}$
b88b
$$
\frac{+121}{a 00 a}
$$

Therefore, in either case, there is only one possible value of $C$, namely 121. An important step is to check whether it is actually possible to find values of $A$ and $B$ which give $C=121$, and working on from one of the above cases, it is easy to find an example, such as $2112-1991=121$.