ABCDE x 4 = EDCBA?

Question

ABCDE × 4 = EDCBA.
Solve for A,B,C,D, and E where each is a unique integer from 0 to 9.

Answer 1

Given:
ABCDE
x 4
= EDCBA

The answer is:
21978
x 4
= 87912

But rather than just give you the answer, here's how I figured it out. First, it is obvious that A must be an even number, because we are multiplying by 4 (an even number). The last digit will therefore be even. It can't be 0, because that would make ABCDE a four-digit number. It can't be more than 2, because that would result in a six-digit answer. So A = 2.

2BCDE
x 4
= EDCB2

So what can E be? The choices are E = {3, 8} because 3 x 4 = 12 and 8 x 4 = 32. But a value of 3 doesn't work in the result (3????) because it is too small.

2BCD8
x 4
= 8DCB2

Since the final number is 8 and we have 2 x 4, that means there is no carry from the prior multiplication (4 x B + carry). So B can't be anything higher than 1, possibly 0.

Looking at the other side of the equation, we have 4D + 3 = (a number ending in 0 or 1). In other words, 4D must end in 7 or 8. Obviously only 8 works, because 4 is an even number. Working forward again, that means B = 1.

21CD8
x 4
= 8DC12

So what values of 4D result in a number ending 8? 4 x 2 = 8, 4 x 7 = 28. Now 2 is already taken and the problem said the digits were unique. So D = 7.

21C78
x 4
= 87C12

Finally, we have a carry of 3 (from 28 + 3 = 31). And when we calculate 4C + 3 it must also result in a carry of 3 and a last digit of C. In other words:
4C + 3 = 30 + C

This is easy to solve:
3C = 27
C = 9

Thus the final answer is:
21978
x 4
= 87912

Answer 2

1

Answer 3

A must be 1 or 2 since any other number would means that the answer would have 6 digit. A also have to be even since 4 time any number is even. thus A is 2. E would have to be 3 or 8. but E cannot be 3 since the last digit have to be 8 or 9. thus E is 8. now 4xB must be less than 10 thus B is one or 2. but A is already 2 thus B is one. 4 x D +3 must end with 1 two possible digit 2 and 7 but 2 already used thus D is 7. 4xC+2 must end with C. two possible numbers 0 and 9. but 0 is not an option thus c=9
so the answer is A=2, B=1, C=9, D=7 and E is 8

Answer 4

A = 2
B = 1
C = 9
D = 7
E = 8

A has to be 2 by because it has to be even and smaller than 3
Thus E has to be 8.

Then solve for:
4D + 3 = B + 10Y, where Y is an element of the set S = {0,1,2,3,4,5,6,7,8,9}.
4C + Y = C + 10Z, where Z is in the set S.
4B + Z = D. .

Answer 5

The solution is ABCDE=21978, but I don't have an analytical approach for obtaining it efficiently. Certainly there's no way A can be greater than 2.

Answer 6

Es sencillo; 0 no es porque sería una cifra de 4 digitos,
1 no puede ser porque no es multiplo de 4
ejemplo-, 1xxxx no puede ser xxxx1
3 no porque multiplicado por 4 se pasa de digitos
es 2 la primera cifra y ahi comenzamos:

2xxxx por 4 = xxxx2

2xxxx
2xxxx
2xxxx
2xxxx
--------
8xxxx de esta forma ya tenemos el ultimo numero

2Bxx8
2Bxx8
2Bxx8
2Bxx8
-------

Answer 7

i think information provided here is not enough.. to solve.

Answer 8

No. I refuse.