# Thread: [SOLVED] What is the 4-digit number?

1. ## [SOLVED] What is the 4-digit number?

If ABCD x A= DCBA, what are the four digit numbers that give you this? And what is the equation? Without repitition?

2. Hello, MSYORK89!

Is there a typo?

If $\displaystyle ABCD \times A\:=\: DCBA$, what are the four-digit numbers?
We have:

. . $\displaystyle \begin{array}{cccc} 1 & 2 & 3 & 4 \\ \hline A & B & C & D \\ \times & & & A \\ \hline D & C & B & A \end{array}$

If $\displaystyle A =0$, the product is zero. .Hence, $\displaystyle A \neq 0.$

If $\displaystyle A = 1$, the product is $\displaystyle ABCD.$ .Hence, $\displaystyle A \neq 1.$

If $\displaystyle A \geq 4$, the product is a five-digit number. .Hence, $\displaystyle A \:=\:2\text{ or }3$

If $\displaystyle A = 2$, we have:

. . $\displaystyle \begin{array}{cccc} 1 & 2 & 3 & 4 \\ \hline 2 & B & C & D \\ \times & & & 2 \\ \hline D & C & B & 2 \end{array}$

In column-4, $\displaystyle D \times 2$ ends in 2.
. . Hence, $\displaystyle D \,=\,1\text{ or }6$

But in column-1, we see that $\displaystyle D\,=\,4\text{ or }5$

. . Therefore: .$\displaystyle A \,\neq\,2$

If $\displaystyle A = 3$, we have:

. . $\displaystyle \begin{array}{cccc} 1 & 2 & 3 & 4 \\ \hline 3 & B & C & D \\ \times & & & 3 \\ \hline D & C & B & 3 \end{array}$

In column-4, we see that: .$\displaystyle D = 1$

But in column-1, $\displaystyle D \,=\,9$

. . Therefore: .$\displaystyle A \,\neq\,3$

The problem has no solutions.

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If the problem is: .$\displaystyle ABCD \times {\color{red}4} \:=\:DCBA$

. . there is a solution: .$\displaystyle 2178 \times 4 \:=\:8712$