Given that A,B,C,D,E each represent one of the digits between?

Question

Given that A,B,C,D,E each represent one of the digits between
1 and
9 and that the following multiplication holds:
A B C D E
X 4
--------------
E D C B A
--------------

what digit does E represent ?

a) 4
b) 6
c) 8
d) 7

please explain the procedure

please give a very detailed answer....

Answer 1

since there's only 5 digits at the answer, and A is the second number from E x 4, so A must be 2.

since E x 4 = _ 2 and E is more or equal 8, so E must be 8.

since B x 4 is less than 10, than B must be 1

since D x 4 + 3 = _1, yhan D must be 7

since C x 4 must be more than 30, C must be 6 or 9 (7&8 already been used). and with a little trial C = 9.

A = 2
B = 1
C = 9
D = 7
E = 8.
Prove = 21978 x 4 = 87912

so the answer is c) 8

if my answer's right, please choose it as best answer =)

Answer 2

The product, E * 4, is always even for any value of E.
Therefore, A must be an even digit, either 2, 4, 6 or 8.

The product, A * 4 + carry = E, means that A can only be 1 or 2.
But A must be even, as shown. Therefore, A = 2.

With A = 2, A * 4 + carry = 8 + carry, so E = 8 or 9.

But E * 4 = tens digit + A = tens digit + 2, so E is either 3 or 8, because only 3 * 4 and 8 * 4 end in a 2.

From both the last statements, E must be 8.

Answer 3

4* E = A, and since the product has as many digits as ABCDE, A < 3 (since 4*3 > 10 but 4*2 < 10)...therefore E * 4 = 1 or 2, but the only possibility is E*4 = 2 (since 1 is odd), and from the list above, E then = 6. (4*6 = 20 + 4)

Steve

Answer 4

i trust that E is going to equivalent 8. in case you do 4*E the most suitable digit of this ought to equivalent A. the purely one which works is 8. "A" won't be able to be more effective than 9 because the answer remains decrease than 99999, hence 4*A won't be able to be more effective than 10 which limits "A" to at least a million or 2. bear in mind that the first 4*E does not ought to equivalent a unmarried digit.