`ABCxxDEED=ABCABC` find `D+E` = ?

Answer

To solve the equation \( ABC \times DEED = ABCABC \), we can follow these steps:

### Step 1: Understand the structure of the numbers
The number \( ABC \) represents a three-digit number, where \( A, B, \) and \( C \) are its digits. The number \( DEED \) represents a four-digit number, where \( D \) and \( E \) are its digits. The number \( ABCABC \) is formed by repeating the three-digit number \( ABC \) twice.

### Step 2: Express \( DEED \) in terms of \( ABC \)
The number \( DEED \) can be expressed as:
\[ DEED = 1000D + 100E + 10E + D = 1001D + 110E \]

### Step 3: Express \( ABCABC \) in terms of \( ABC \)
The number \( ABCABC \) can be expressed as:
\[ ABCABC = 1000 \times ABC + ABC = 1001 \times ABC \]

### Step 4: Set up the equation
From the problem statement, we have:
\[ ABC \times (1001D + 110E) = 1001 \times ABC \]

### Step 5: Simplify the equation
Assuming \( ABC \neq 0 \) (since it's a three-digit number), we can divide both sides by \( ABC \):
\[ 1001D + 110E = 1001 \]

### Step 6: Rearrange the equation
Rearranging gives us:
\[ 1001D + 110E - 1001 = 0 \]

### Step 7: Solve for \( D \) and \( E \)
This can be rewritten as:
\[ 1001D + 110E = 1001 \]
To find integer values for \( D \) and \( E \), we can try different values for \( D \) (from 0 to 9) and solve for \( E \).

1. If \( D = 1 \):
\[ 1001(1) + 110E = 1001 \implies 110E = 0 \implies E = 0 \]

Thus, we have \( D = 1 \) and \( E = 0 \).

### Step 8: Calculate \( D + E \)
Now, we can find \( D + E \):
\[ D + E = 1 + 0 = 1 \]

### Final Answer
Thus, the final answer is:
\[ D + E = 1 \]

---