HongKong 2022-23 IMO Selection Tests

Answer: $2358$

Note that $m$ is of the form $ABCBA$ with $A$ nonzero, so $n$ is of the form $BCBA$, $ACBA$, $ABBA$, $ABCA$ or $ABCB$. The second, third and fourth cases can be combined, so we are down to three types of possible values of $n$:

Type I — Equal thousands digit and tens digit, with nonzero unit digit;

Type II — Equal thousands digit and unit digit;

Type III — Equal hundreds digit and unit digit.

There are $9\times 10\times 9=810$ possibilities of Type I (9 choices for the common thousands and tens digit, 10 choices for the hundreds digit and 9 choices for the unit digit), and similarly $9\times 10\times 10=900$ possibilities for each of Type II and Type III. However,

90 numbers are of both Types I and II (those of the form $XYXX$ with $X$ nonzero);

81 numbers are of both Types I and III (those of the form $XYXY$ with both $X$ and $Y$ nonzero);

90 numbers are of both Types II and III (those of the form $XXYX$ with $X$ nonzero);

* 9 numbers are of all three types ($1111$, $2222$, ..., $9999$).

By the inclusion-exclusion principle, the answer is $810+900+900-90-81-90+9=2358$.