ABCD×E=DCBA (A,B,C,D,E is all different number in 1~9)

Question

Solve $ABCD×E=DCBA$, where $A,B,C,D,E$ are all different numbers from $1-9$

I know the answer is $ABCD=2178$, $E=4$ , so $2178\times4=8712$ but I don't know how can we find this answer.

To solve this problem, I discover the fact that $d>e$ as $dcba=e\times abcd>e\times1000$,so $e\leq d $ but $d$ and $e$ are different so $d>e$ and because $e$ is not 1; (if $e=1$, then $abcd=dcba$) $e>1$ so a is number among $1,2,3,4$.

After this how can I find the answer?

Answer 1

Further note that $a<d$ as $ae<d$. Since $ed\equiv a$ mod $10$ we find that $de>10$. Since $d<e$ we find $d\geq 4$.

• If $d=4$ then $e=3$ and thus $ed=12$ implies $a=2$, but then $ea=6>d$ wich is a contradiction.
• If $d=5$ then $de$ modulo $10$ is either $0$ or $5$, so $a=0$, which is not allowed or $a=5=d$ which is not allowed.
• If $d=6$, then $de$ is even, so $a$ must be even. Since $ae<d$ it follows that $a=2$ and $e=3$. Then $de=18$ so $a=8$ which is a contradiction.
• If $d=7$, then $ae=d$ or $ae+1=d$. If $ae=d$ either $e=7$ or $a=7$ which is a contradiction. So $ae=6$ and we find that $\{a,e\}=\{1,6\}$ or $\{a,e\}=\{2,3\}$. But if $e$ is odd, then $de$ is odd and hence $a$ is odd and if $e$ is even, then $de$ is even and hence $a$ is even.
• If $d=9$, then $ae=9$ or $ae=8$. If $ae=9$ then $a=e=3$ which is not allowed so $ae=8$ and $\{a,e\}=\{2,4\}$. If $e=2$ we find that $de=18$ so $a=8$, if $e=4$ then $de=36$ so $a=6$, both give a contradiction.

So $d=8$. We find that $ae=8$ or $ae=7$. If $ae=7$, then $a=1$ and $e=7$, but $de$ is even, so $a$ is even, which is a contradiction. So $ae=\{2,4\}$. If $e=2$ then $de=16$, so $a=6$, which is a contradiction. Hence $e=4$ and $a=2$. It also follows that $be<10$, so $b=1$ or $b=2$. Since $a=2$ we find that $b=1$.

Finally we have, since $de=32$, that $1=b\equiv ce+3=4c+3$ mod $10$. So $c=2$ or $c=7$, which by the above means that $a=2$, $b=1$, $c=7$, $d=8$ and $e=4$.

There might be a more elegant way of obtaining this.