MHB 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&amp;_2&amp;_3&amp;_4 \\<br /> A&amp;B&amp;C&amp;D \\<br /> \times &amp;&amp;&amp; 4 \\<br /> \hline D&amp;C&amp;B&amp;A<br /> \end{array}

In column-1: \;4\!\cdot\!A\text{ (plus &#039;carry&#039;)} \,=\, 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&amp;_2&amp;_3&amp;_4 \\<br /> 2&amp;B&amp;C&amp;8 \\<br /> \times &amp;&amp;&amp; 4 \\<br /> \hline 8&amp;C&amp;B&amp;2<br /> \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}<br /> 2&amp;1&amp;7&amp;8 \\<br /> \times &amp;&amp;&amp; 4 \\<br /> \hline 8&amp;7&amp;1&amp;2<br /> \end{array}

 

[anemone]

Hello, anemone!


We have: \;\;\begin{array}{cccc}_1&amp;_2&amp;_3&amp;_4 \\<br /> A&amp;B&amp;C&amp;D \\<br /> \times &amp;&amp;&amp; 4 \\<br /> \hline D&amp;C&amp;B&amp;A<br /> \end{array}

In column-1: \;4\!\cdot\!A\text{ (plus &#039;carry&#039;)} \,=\, 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&amp;_2&amp;_3&amp;_4 \\<br /> 2&amp;B&amp;C&amp;8 \\<br /> \times &amp;&amp;&amp; 4 \\<br /> \hline 8&amp;C&amp;B&amp;2<br /> \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}<br /> 2&amp;1&amp;7&amp;8 \\<br /> \times &amp;&amp;&amp; 4 \\<br /> \hline 8&amp;7&amp;1&amp;2<br /> \end{array}

Good job, soroban! (Yes)