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 [tex]ABCD[/tex] which, when multiplied by 4,
give a product equal to the number with the digits reversed, $DCBA$.

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

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

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

And it follow that [tex]D = 8.[/tex]

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

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

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

Therefore:[tex]\;\begin{array}{cccc}
2&1&7&8 \\
\times &&& 4 \\
\hline 8&7&1&2
\end{array}[/tex]

 

[anemone]

Hello, anemone!


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

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

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

And it follow that [tex]D = 8.[/tex]

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

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

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

Therefore:[tex]\;\begin{array}{cccc}
2&1&7&8 \\
\times &&& 4 \\
\hline 8&7&1&2
\end{array}[/tex]

Good job, soroban! (Yes)