Question

I found an interesting problem:

ABCD x E = DCBA

so each letter is really a digit, and they're all different digits. So it could be ABCD=1234 or something like that.

Any ideas on how to solve this thing?

Answer 1

What we have to find

Answer 2

@Kainui

Answer 3

What are the numbers that make this a true statement is all.

Answer 4

so if A is 2 then B should be 3 for example

Answer 5

?

Answer 6

Here's my attempt, it's obviously wrong but fun to look at. \[(A*10^3+B*10^2+C*10+D)*E=D*10^3+C*10^2+B*10+A\]\[AE*10^3+BE*10^2+CE*10+DE=D*10^3+C*10^2+B*10+A\] Collect all the terms with the same power of 10 (obviously wrong but follow along)

AE=D, BE=C, CE=B, and DE=A

take the log of all these equations to get:

logA+logE-logD=0, logB+logE-logC=0, etc... then plug them into a matrix:
\[\left[\begin{matrix}1 & 0&0&1&-1 \\ 0 & 1&-1&0&1 \\0 & -1&1&0&1 \\-1 & 0&0&1&1 \end{matrix}\right]\left(\begin{matrix}logA \\ logB\\logC\\logD\\logE\end{matrix}\right)=\left(\begin{matrix}0 \\ 0\\0\\0\end{matrix}\right)\]

row reduce the augmented matrix and you solve to get

A=D, B=C, and E=1. So obviously this is wrong because we can't have any of the digits equal to each other, but it works if that wasn't a constraint.

ABBA*1=ABBA

So that's the best I've got lol.

Answer 7

A, B, C, D, and E are all just digits between 1 and 9 and none of them are equal. They can be anything as long as they make the equation work out. So for instance, if A=2 and B=4 and C=5 then AB x C = 24*5=96

Answer 8

i think that works only if there was no carry any where in the multiplication by E

Answer 9

Yeah the problem is the carrying. So I could either change base or try to figure out how to use the modulus function maybe? Idk.

DEmod10=A
CEmod100=B
etc...

I think that's how it works lol idk.

Answer 10

Assuming A, B, C, D and E are distinct digits, then we can say that:

E > 1

and therefore that:
A < 5

Answer 11

what do u mean by E ?

Answer 12

if A=1 then D*E must be a product that ends in a "1", the only combinations are:

7x3 OR 3x7

which means we either have:
1BC7 x 3 = 7CB1 (invalid combination)

or:
1BC3 x 7 = 3CB1 (cannot be possible)

therefore A > 1 and therefore E < 5

so now we know that:
1 < A < 5
1 < E < 5

Answer 13

now we need to find combinations of DxE that yield A as their units digit

A=2 => DxE = 4x3 OR 3x4 OR 4x8 OR 8x4
A=3 => DxE = no valid combinations
A=4 => DxE = 2x7 OR 7x2 OR 3x8 OR 8x3

Answer 14

E is just another number. An example of a wrong solution would be:

2583*7=3852.

Answer 15

which narrows A down to either 2 OR 4

Answer 16

Working through this reduced set I get one valid solution:

2178 x 4 = 8712

Answer 17

I don't see why A can't be 1, 2, 3, or 4.

Answer 18

do you agree that if A=1 then D*E must be a product that ends in a "1"?

Answer 19

(D+10C+100B+1000A)E=A+10B+100C+1000D
D*E=A
C*E=B
B*E=C
A*E=D

Answer 20

so it work if :-
A/D=B/C
i got a ratio :3
lets make sure of it
(so we must have no zero )

Answer 21

ohh we should know E :-|

Answer 22

@BSwan - you are assuming there will be no carry

Answer 23

Here are some equations I noticed:

DE mod10 = A
DE mod100 - DE mod10 + CE mod10 = B
CE mod100 - CE mod10 + BE mod10 = C
BE mod100 - BE mod10 + AE mod10 = D
AE mod100 - AE mod10 = 0

These all come from making sure each digit lines up. Noticing a couple extra things, since we don't have repeating digits,

E=/=1

For any situation the highest possible combination of digits can be 9*8, which is 72. So this means that if X and Y are any digits, then

XY mod100 - XY mod10 <= 7

I combined all first equations at the top and got:

A + B + C + D = AE mod100 + BE mod100 + CE mod100 + DE mod100

then using that last inequality I came up with:

A + B + C + D <= 4*7 + AE mod10 + BE mod10 + CE mod10 + DE mod10

From here I have absolutely no clue if this is in the right direction or not because I really don't know how to use modulus with algebra.