Simultaneous verbal arithmetic

Question

1. A B C D + A E F = E F G H and
2. I C J * E = A B C D

Each letter represents a unique digit from 0 to 9 and both equations must hold true.

Having some trouble solving this, here is what I have so far:

B + A = 10 or B + A > 10 since something must be carried over to the thousands column (A) in equation 1

I also know that A + 1 = E since the 1 is carried over

17 >= B + A >= 10

A cannot equal 0,9

Any idea how to solve this? Any contributions are greatly appreciated.

Answer 1

Putting this up. If anyone has partial insights, feel free to add to this post instead of commenting (since writing as is text is difficult in the comments).

So far, we have

  A    B  C    D
+      A (A+1) F
--------------
 (A+1) F  G    H

   I  C  J
x       (A+1)
-----------
A  B  C  D

Answer 2

A quick computer search finds the unique answer:

$A, B, \dots, J = 6, 5, 3, 8, 7, 2, 1, 0, 9, 4$.

Hence there is not much to explain. Without any context, I cannot tell whether it is intended to solve it by computer or without computer.

Answer 3

Well, it's doable by hand. At least, I did it (with a couple of sheets of paper...)

We know that A + 1 = E. Look now at xCJ * E = xxCD and search for all the possible combinations for (A) CJED: they are

A C J E D 
1 9 7 2 4
1 9 8 2 6
2 4 7 3 1 
2 4 9 3 7 
2 9 7 3 1
2 9 8 3 4 
3 6 7 4 8 
3 9 8 4 2 
5 3 9 6 4
5 7 9 6 4 
6 3 4 7 8
6 8 3 7 1

Moreover, from the multiplication we have a range of the possible values of I:

A E  I
1 2 5-9
2 3 6-9
3 4 7-9
5 6 8-9
6 7 8-9

Putting all together and completing the multiplication, we have

A C J E D I B
1 9 7 2 4 6 3
1 9 8 2 6 7 5
2 4 7 3 1 8 5
2 4 7 3 1 9 8
2 4 9 3 7 8 5
2 9 7 3 1 6 0
2 9 7 3 1 8 6
2 9 8 3 4 6 0
3 9 8 4 2 7 1
5 3 9 6 4 8 0
5 7 9 6 4 8 2
6 3 4 7 8 9 5

From here, considering D + F = (1)H we remain with

A B C D E F G H I J
1 5 9 6 2 4   0 7 8
2 0 9 1 3 4   5 6 7
2 6 9 1 3 4   5 8 7
2 0 9 4 3 1   5 6 8
6 5 3 8 7 2   0 9 4

and only the last line (with G = 1) satisfies the sum.