1
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  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.

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  • 1
    $\begingroup$ You also know that $E = A + 1$, because the carried digit must have been a $1$. $\endgroup$ Jul 14 '20 at 15:21
  • $\begingroup$ @Omnomnomnom Yeah sorry, forgot to add that to my question. Thanks $\endgroup$ Jul 14 '20 at 15:22
1
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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.

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4
  • $\begingroup$ without a computer preferably $\endgroup$ Jul 14 '20 at 15:48
  • $\begingroup$ No idea for the moment. Sometimes these puzzles are even generated with the help of computers in the first place, so I'm not really willing to spend time thinking about a no-computer solution... $\endgroup$
    – WhatsUp
    Jul 14 '20 at 18:17
  • $\begingroup$ I wrote a program to go through all possibilities of $I, C, J, E$. This then determines $A, B, C, D$. I then go through remaining possibilities of $F$, which determines $G$ and $H$. $\endgroup$
    – WhatsUp
    Jul 14 '20 at 18:35
  • $\begingroup$ It was in C, and I already deleted it... It's not that tricky and you may write it yourself, with Python or any of your preferred programming languages. $\endgroup$
    – WhatsUp
    Jul 15 '20 at 12:15
1
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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.

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0
$\begingroup$

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
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