Four digit reversal numbers How to prove without an exhaustive checking that there are only 2 (nontrivial) four digit reversal numbers?
 A: That depends how exhaustive you are willing to be.  The multiplier can only be in the range $2-9$, which is only eight cases.  To check $4$ we write $abcd \times 4=dcba$ where concatenation indicates different digits.  $a$ has to be even as the last digit of the product, and has to be $2$ or there would be a carry.  Then $d$ has to be $8$ so that it can produce $a$.  $b$ has to be $0,1,2$ to avoid a carry and is odd because of the carry from the ones place, so it is $1$.  Then $c$ has to be $7$ and we are done-$2178 \times 4 = 8712$.  $5, 6$ and $8$ are out because $a$ would have to be $1$  to avoid a product over $10,000$ but that is odd and not $5$.  $7$ is out because again $a$ would have to be $1$, but then $d$ can't be $3$.  So we just have to check $2, 3,$ and $9$-not too much work.
A: Let $A = \{2,3,4,5,6,7,8,9\}$.
Let the $4$ digit reversal number be $abcd$ i.e. $1000a + 100b + 10 c + d$, where $a \in \{1\} \cup A$, $d \in \{1,2,3,\ldots,a-1\}$ and $b,c \in \{0,1\} \cup A$. Its reversed number is $$dcba = 1000d + 100c + 10b + a$$ We want $abcd = k \times (dcba)$ where $k \in A$. This gives us
$$(1000-k)a + (100-10k)b + (10-100k)c + (1-1000k) d = 0$$
$$\color{red}{\text{First note that $10 \vert (ka-d)$. This also means that if $a$ is even, then $d$ also has to be even.}}$$
Let us call the above necessary condition in red as $\star$. As explained below, this necessary condition filters out almost all the solutions that are not possible.


*

*If $a=2$. Then $d=1$. Violates $\star$.

*If $a=3$. Then $d=1$. Hence, $k=1$ or $2$. $ka - d = 1 or 5$. Violates $\star$.

*If $a=4$. Then $d=2$. If $d=2$, then $k=2$, then $ka-d = 6$. Violates $\star$.

*If $a=5$. Then $d=1$ or $2$. If $d=1$, $k = 3$ or $4$ or $5$. But $10$ doesn't divide $ka - d$ if $d=1$. Violates $\star$. If $d=2$, $k=2$. Violates $\star$.

*If $a=6$, then $d=2$. If $d=2$, then $k=3$.  Violates $\star$.

*If $a=7$, then $d=1,2,3$. If $d=1$, $k=4,5,6,7$.  Violates $\star$. If $d=2$, then $k=3$.  Violates $\star$.

*If $a=8$, then $d=2,4$. If $d=4$, $k=2$. Violates $\star$. If $d=2$, $k=3,4$. $k=3$ violates $\star$. However, $k=4$ satisfies $\star$. Hence, $a=8$, $d=2$ and $k=4$ is a possible candidate. Let us return to this candidate later.

*If $a=9$, then $d=1,2,3,4$. If $d=1$, $k=5,6,7,8,9$. Except $k=9$, the rest violate $\star$. We will return to the case $a=9$, $d = 1$ and $k=9$ later. If $d=2$, then $k=4$. Violates $\star$. If $d=3$, $k=3$. Violates $\star$. If $d=4$, $k=2$. Violates $\star$.


Hence, after this elimination using the necessary condition $\star$, we get only two possibilities.


*

*$a = 8$, $d=2$ and $k=4$. This means $60b - 390c + 996 \times 8 - 3999 \times 2 = 0$. Hence, $2b - 13c = 1$. This gives $b=7$ and $c=1$.

*$a = 9$, $d=1$ and $k=9$. This means $10b - 890c + 991 \times 9 - 8999 \times 1$. Hence, $b-89c = 8$. This gives us $b=8$ and $c=0$.


Hence, the only two numbers are $8712$ and $9801$.
