Help me solve the equation 4xABCD=DCBA.?

Question

How do you solve the equation 4*ABCD=DCBA? ABCD is a number containing the whole numbers A, B, C and D.

Can you figure out which numbers A,B,C and D are?

Maybe I was a bit unclear in my question, but Green G gave the correct answer (great answer). The problem is that I suck at programming, so if anyone can solve it without programming and without "testing" each number it would be great.

Answer 1

A has to be <3 because 3*4 = 12 so RHS shall be 5 digit number.
A cannor be 1 as from RHS A has to be even

so A has to be 2
now B can be either 1 or 3 or 5 or 7 but B < 5 because 4*25 = 100 that is 5 digit.

So AB = 21 or 23
if AB = 23 DC >= 92
do 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
60C = 420 C =7
so number = 2178*4 = 8712

Answer 2

I know!

Based on observation, the answer should be a 4 digit number...
therefore, the value of A could either be 1 or 2. with this information we could say that D could either be 4 or 8
we then solve...

ABCD * 4 = DCBA

with 1 & 4

1BC4 * 4 = 4CB1

we could see that 4 * 4 is not equal to a number that has a ones digit of 1...

with 2 & 8

2BC8 * 4 = 8CB2

the answer to 4 * 8 has a ones digit of 2 so the value of A & D repectively is 2 & 8

For B & C

if we analyze the question carefully we would arrive with these equations and inequalities...

4C + 3 = B + 10x ( x is the number to be carried )
4B + x = C
B < 10
C < 10
B & C are not equal to 2 and 8

based on alaysis... B can not be greater than 3 because if it is the value of D & A would change and if A is Already 2 we are left with B = 1

subtituting to the first two equations I've mentioned earlier...
4C + 3 = B + 10x & 4B + x = C

4C + 3 = 1 + 10x & 4 + x = C

4C + 2 = 10x & 4 + x = C

WE THEN SOLVE THEM SIMULTANEOUSLY!!!

4 ( 4 + x ) + 2 = 10x
16 + 4x + 2 = 10x
18 = 6x
3 = x

4 + x = C
4+ 3 = 7

In conclusion...

ABCD is 2178
and
DCBA is 8712

Answer 3

The only one with the correct answer, so far is denxxchua. The only number you can multiply by four to equal itself is zero. SO ABCD has to equal 0 if 4 (ABCD) = ABC. [4 x0 = 0].

If the product of A*B*C*D = 0. Then A, B, C, and D must all equal 0.

("Denxxchua" was first with the correct answer, so he is probably your "best answer")

Answer 4

Since the variables on the left side of the equation are being multiplied by 4 and you have all of the same variables on the right side of the equation, you must multiply everything on the right by 4.

4*ABCD= DCBA*4

Answer 5

4*A = D implies that 4 is a divisor of D thus D = 4 OR D=8 OR D=0

since there is no carry in any cases :
4*C = B => B=4OR B=8 B=0

4*B = C => C=4OR C=8 C=0

4*A = D => D=4OR D=8 D=0

the continuation is trivial

Answer 6

A=0
B=0
C=0
D=0

Answer 7

All the letters are equal to 1. !*1*1*1

Answer 8

Let x=0
Let A,B,C or D = x

Sub x into equation.

Therefor A, B, C and D equals to 0.

Answer 9

2178

4 * 2178 = 8712

Here's a little Bourne shell I cobbled together for all you Linux types:

#!/bin/sh

x="1000"
A=`echo $x | cut -c1`
B=`echo $x | cut -c2`
C=`echo $x | cut -c3`
D=`echo $x | cut -c4`

while [ $x -lt 10000 ]; do
y=`expr $x \* 4`
if [ "$y" = "${D}${C}${B}${A}" ]; then
echo $y
fi
let x+=1
A=`echo $x | cut -c1`
B=`echo $x | cut -c2`
C=`echo $x | cut -c3`
D=`echo $x | cut -c4`
done

exit 0

Answer 10

Lil Tatoo says, "D Cab boss.... D Cab!"