1. In the following addition problem find $$ \mathbf{A}+\mathbf{B}+\mathbf{C} $$ if each digit is different $$ \begin{array}{c} A B C \ \text { ABC } \ \text { ABC } \ \hline 1 \text { A } 25 \end{array} $$

Question

UpStudy AI · Accepted Answer

To solve the addition problem, we need to analyze the given equation:

$$
\begin{array}{c}
A B C \
\text { ABC } \
\text { ABC } \
\hline 1 \text { A } 25
\end{array}
$$

This can be interpreted as:

$$
ABC + ABC + ABC = 1A25
$$

This simplifies to:

$$
3 \times ABC = 1A25
$$

Let $$ ABC $$ be represented as $$ 100A + 10B + C $$. Therefore, we can rewrite the equation as:

$$
3(100A + 10B + C) = 1000 + 100A + 20 + 5
$$

This simplifies to:

$$
300A + 30B + 3C = 1000 + 100A + 25
$$

Now, rearranging gives us:

$$
300A + 30B + 3C - 100A - 25 = 1000
$$

This simplifies to:

$$
200A + 30B + 3C = 1000 + 25
$$

Thus:

$$
200A + 30B + 3C = 1025
$$

Now, we can divide the entire equation by 5 to simplify:

$$
40A + 6B + C = 205
$$

Next, we need to find values for $$ A $$, $$ B $$, and $$ C $$ such that they are different digits (0-9) and satisfy the equation $$ 40A + 6B + C = 205 $$.

### Step 1: Determine possible values for $$ A $$

Since $$ A $$ is a digit, the maximum value for $$ 40A $$ must be less than or equal to 205. Thus, we can find the maximum possible value for $$ A $$:

$$
40A \leq 205 \implies A \leq \frac{205}{40} \approx 5.125
$$

So, $$ A $$ can be 0, 1, 2, 3, 4, or 5.

### Step 2: Test values for $$ A $$

Let's test each possible value for $$ A $$:

1. **If $$ A = 5 $$**:
   $$
   40(5) + 6B + C = 205 \implies 200 + 6B + C = 205 \implies 6B + C = 5
   $$
   This is not possible since $$ B $$ and $$ C $$ must be digits.

2. **If $$ A = 4 $$**:
   $$
   40(4) + 6B + C = 205 \implies 160 + 6B + C = 205 \implies 6B + C = 45
   $$
   This is not possible since $$ C $$ must be a digit.

3. **If $$ A = 3 $$**:
   $$
   40(3) + 6B + C = 205 \implies 120 + 6B + C = 205 \implies 6B + C = 85
   $$
   This is not possible since $$ C $$ must be a digit.

4. **If $$ A = 2 $$**:
   $$
   40(2) + 6B + C = 205 \implies 80 + 6B + C = 205 \implies 6B + C = 125
   $$
   This is not possible since $$ C $$ must be a digit.

5. **If $$ A = 1 $$**:
   $$
   40(1) + 6B + C = 205 \implies 40 + 6B + C = 205 \implies 6B + C = 165
   $$
   This is not possible since $$ C $$ must be a digit.

6. **If $$ A = 0 $$**:
   $$
   40(0) + 6B + C = 205 \implies 6B + C = 205
   $$
   This is not possible since $$ C $$ must be a digit.

### Conclusion

After testing all possible values for $$ A $$, we find that there are no valid combinations of $$ A $$, $$ B $$, and $$ C $$ that satisfy the equation while keeping all digits different.

Thus, it appears there may be an error in the problem setup or constraints. Please verify the problem statement or provide additional information.
There are no valid digits $$ A $$, $$ B $$, and $$ C $$ that satisfy the equation $$ 3 \times ABC = 1A25 $$ with all digits different.

1. In the following addition problem find \( \mathbf{A}+\mathbf{B}+\mathbf{C} \) if each digit is different \[ \begin{array}{c} A B C \\ \text { ABC } \\ \text { ABC } \\ \hline 1 \text { A } 25 \end{array} \]

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