Find al the four-digit numbers ABCD

[anemone]

Find all the four-digit numbers $ABCD$ which when multiplied by $4$ give a product equal to the number with the digits reversed, $DCBA$. (The digits do not need to be different.)

 

[kaliprasad]

Find all the four-digit numbers $ABCD$ which when multiplied by $4$ give a product equal to the number with the digits reversed, $DCBA$. (The digits do not need to be different.)

Spoiler

A has to be < 3 because 3* 4 = 12 so RHS is a 5 digit number
A cannot be 1 as from RHS A has to be even.
So A has to be 2.
now $B$ can be an odd digit $B \lt 5$ because $4*25 = 100$ that is 5 digit
So $AB = 21 / 23$
if $AB = 23$ $DC \ge 92$ so $D = 9$ which is not possible as $4*8$ is $2$ ending but $4*9$ is not.
$AB = 21$
So $D = 8$
so the number = $4*(2108+10C) = 8032+100C$
or $8432+40C = 8012+ 100C$
or $60C = 420$
so $C =7$
so number = $2178*4 = 8712$ or $ABCD=2178$

 

[anemone]

Well done Kali! And thanks for participating!

 

[soroban]

Hello, anemone!

Find all the four-digit numbers $$ABCD$$ which, when multiplied by 4,
give a product equal to the number with the digits reversed, $DCBA$.

We have: $$\;\;\begin{array}{cccc}_1&_2&_3&_4 \\ A&B&C&D \\ \times &&& 4 \\ \hline D&C&B&A \end{array}$$

In column-1: $$\;4\!\cdot\!A\text{ (plus 'carry')} \,=\, D$$
. . Hence, $$A = 1\text{ or }2.$$

In column-4: $$\;4\!\cdot\!D\text{ ends in }A,$$ an even digit.
. . Hence, $$A =2.$$

And it follow that $$D = 8.$$

We have: $$\;\;\begin{array}{cccc}_1&_2&_3&_4 \\ 2&B&C&8 \\ \times &&& 4 \\ \hline 8&C&B&2 \end{array}$$

There is no 'carry' from column-2.
. Hence, $$B =0\text{ or }1.$$

In column-3, $$4\!\cdot\!C + 3\text{ ends in }B,\text{ an odd digit.}$$
. Hence, $$B = 1\text{ and }C =7.$$

Therefore:$$\;\begin{array}{cccc} 2&1&7&8 \\ \times &&& 4 \\ \hline 8&7&1&2 \end{array}$$

 

[anemone]

Hello, anemone!


We have: $$\;\;\begin{array}{cccc}_1&_2&_3&_4 \\ A&B&C&D \\ \times &&& 4 \\ \hline D&C&B&A \end{array}$$

In column-1: $$\;4\!\cdot\!A\text{ (plus 'carry')} \,=\, D$$
. . Hence, $$A = 1\text{ or }2.$$

In column-4: $$\;4\!\cdot\!D\text{ ends in }A,$$ an even digit.
. . Hence, $$A =2.$$

And it follow that $$D = 8.$$

We have: $$\;\;\begin{array}{cccc}_1&_2&_3&_4 \\ 2&B&C&8 \\ \times &&& 4 \\ \hline 8&C&B&2 \end{array}$$

There is no 'carry' from column-2.
. Hence, $$B =0\text{ or }1.$$

In column-3, $$4\!\cdot\!C + 3\text{ ends in }B,\text{ an odd digit.}$$
. Hence, $$B = 1\text{ and }C =7.$$

Therefore:$$\;\begin{array}{cccc} 2&1&7&8 \\ \times &&& 4 \\ \hline 8&7&1&2 \end{array}$$

Good job, soroban! (Yes)