N is a two - digits number having sum of the digits as S and porduct of digits as P . How many such two -digit numbers exist such that `2N = 2S + 3P`?

Options

A. 0
B. 1
C. 9
D. 12

Answer

To solve the problem, we need to find how many two-digit numbers \( N \) satisfy the equation \( 2N = 2S + 3P \), where \( S \) is the sum of the digits and \( P \) is the product of the digits of \( N \).

### Step-by-step Solution:

1. **Define the two-digit number**:
Let \( N \) be a two-digit number represented as \( N = 10x + y \), where \( x \) is the tens digit and \( y \) is the units digit. Here, \( x \) can take values from 1 to 9 (since \( N \) is a two-digit number) and \( y \) can take values from 0 to 9.

2. **Calculate the sum and product of the digits**:
- The sum of the digits \( S = x + y \)
- The product of the digits \( P = x \cdot y \)

3. **Substitute into the given equation**:
We need to substitute \( N \), \( S \), and \( P \) into the equation:
\[ 2N = 2S + 3P \]
Substituting the values we defined:
\[ 2(10x + y) = 2(x + y) + 3(xy) \]

4. **Simplify the equation**:
Expanding both sides:
\[ 20x + 2y = 2x + 2y + 3xy \]
Now, we can cancel \( 2y \) from both sides:
\[ 20x = 2x + 3xy \]
Rearranging gives:
\[ 20x - 2x = 3xy \]
Simplifying further:
\[ 18x = 3xy \]

5. **Factor out common terms**:
Dividing both sides by 3 (assuming \( x \neq 0 \)):
\[ 6x = xy \]
Rearranging gives:
\[ y = \frac{6x}{x} \quad \text{(as long as \( x \neq 0 \))} \]
Thus:
\[ y = 6 \]

6. **Determine possible values for \( x \)**:
Since \( y = 6 \), we can find the possible values for \( x \). The tens digit \( x \) can take any value from 1 to 9. Therefore, the valid two-digit numbers \( N \) are:
- \( 16 \)
- \( 26 \)
- \( 36 \)
- \( 46 \)
- \( 56 \)
- \( 66 \)
- \( 76 \)
- \( 86 \)
- \( 96 \)

7. **Count the valid two-digit numbers**:
There are 9 valid two-digit numbers that satisfy the condition.

### Final Answer:
The total number of two-digit numbers \( N \) such that \( 2N = 2S + 3P \) is **9**.