Replace A,B and C by suitable numerals. `({:(," "CB5),(,-28A):})/(" "259)`

Answer

To solve the equation `({:(," "CB5),(,-28A):})/(" "259)`, we need to replace A, B, and C with suitable numerals.

Let's break it down step by step:

### Step 1: Understand the structure of the equation
The equation can be interpreted as:
\[ \frac{CB5 - 28A}{259} \]
Here, `CB5` represents a three-digit number where C is the hundreds place, B is the tens place, and 5 is the units place. `28A` represents a three-digit number where 2 is the hundreds place, 8 is the tens place, and A is the units place.

### Step 2: Rewrite the equation
We can express `CB5` and `28A` in terms of their digits:
- `CB5` = \(100C + 10B + 5\)
- `28A` = \(280 + A\)

Thus, the equation becomes:
\[ \frac{(100C + 10B + 5) - (280 + A)}{259} \]

### Step 3: Simplify the equation
Now, let's simplify the numerator:
\[ 100C + 10B + 5 - 280 - A = 100C + 10B - A - 275 \]
So, we have:
\[ \frac{100C + 10B - A - 275}{259} \]

### Step 4: Set up the equation
For the division to be valid, the numerator must be a multiple of 259. Let's denote:
\[ 100C + 10B - A - 275 = k \cdot 259 \]
for some integer k.

### Step 5: Find suitable values for A, B, and C
To find suitable values for A, B, and C, we can start testing values for k. We know that A, B, and C are digits (0-9).

1. **Testing k = 1:**
\[ 100C + 10B - A - 275 = 259 \]
Rearranging gives:
\[ 100C + 10B - A = 534 \]

2. **Testing k = 2:**
\[ 100C + 10B - A - 275 = 518 \]
Rearranging gives:
\[ 100C + 10B - A = 793 \]
This is not possible since \(100C + 10B\) cannot exceed 999.

3. **Testing k = 0:**
\[ 100C + 10B - A - 275 = 0 \]
Rearranging gives:
\[ 100C + 10B - A = 275 \]

### Step 6: Solve for A, B, and C
Now we can solve for A, B, and C:
- Since \(100C + 10B = 275 + A\), we can try different values for A (0-9) and see if we can find suitable values for B and C.

1. **If A = 5:**
\[ 100C + 10B = 275 + 5 = 280 \]
- \(C = 2\) and \(B = 8\) gives us \(CB5 = 285\).

2. **If A = 6:**
\[ 100C + 10B = 275 + 6 = 281 \]
- This does not yield valid digits for B and C.

3. **If A = 4:**
\[ 100C + 10B = 275 + 4 = 279 \]
- \(C = 2\) and \(B = 7\) gives us \(CB5 = 275\).

### Conclusion
After testing various values, we find:
- \(C = 2\)
- \(B = 7\)
- \(A = 5\)

Thus, the suitable numerals are:
- A = 6
- B = 4
- C = 2