If the number in the units place exceeds the number in the tens by 2 in a two-digit number and the product of the number with the sum of its digits is equal to 280 what would be the sum of its digits?

Options

A. 4
B. 9
C. 9
D. 10

Answer

To solve the problem step by step, let's break it down clearly:

### Step 1: Define the digits
Let the unit digit of the two-digit number be \( x \) and the tens digit be \( y \). Therefore, the two-digit number can be represented as:
\[ 10y + x \]

### Step 2: Set up the relationship between the digits
According to the problem, the number in the units place exceeds the number in the tens place by 2. This gives us the equation:
\[ x = y + 2 \]

### Step 3: Set up the equation involving the product of the number and the sum of its digits
The sum of the digits is:
\[ x + y \]
The product of the number and the sum of its digits is given as:
\[ (10y + x)(x + y) = 280 \]

### Step 4: Substitute \( x \) in the product equation
Substituting \( x \) from Step 2 into the product equation:
\[ (10y + (y + 2))((y + 2) + y) = 280 \]
This simplifies to:
\[ (10y + y + 2)(2y + 2) = 280 \]
\[ (11y + 2)(2y + 2) = 280 \]

### Step 5: Expand the equation
Now, expand the left-hand side:
\[ (11y + 2)(2y + 2) = 22y^2 + 22y + 4y + 4 = 22y^2 + 26y + 4 \]
Setting this equal to 280 gives us:
\[ 22y^2 + 26y + 4 = 280 \]

### Step 6: Rearrange the equation
Rearranging the equation:
\[ 22y^2 + 26y + 4 - 280 = 0 \]
\[ 22y^2 + 26y - 276 = 0 \]

### Step 7: Simplify the equation
Dividing the entire equation by 2 for simplicity:
\[ 11y^2 + 13y - 138 = 0 \]

### Step 8: Factor the quadratic equation
Now, we need to factor the quadratic equation. We look for two numbers that multiply to \(-138 \times 11\) and add to \(13\). This can be factored as:
\[ (11y - 6)(y + 23) = 0 \]

### Step 9: Solve for \( y \)
Setting each factor to zero gives:
1. \( 11y - 6 = 0 \) → \( y = \frac{6}{11} \) (not a valid digit)
2. \( y + 23 = 0 \) → \( y = -23 \) (not a valid digit)

However, we can also use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 11, b = 13, c = -138 \):
\[ y = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 11 \cdot (-138)}}{2 \cdot 11} \]
Calculating the discriminant:
\[ y = \frac{-13 \pm \sqrt{169 + 6072}}{22} \]
\[ y = \frac{-13 \pm \sqrt{6241}}{22} \]
\[ y = \frac{-13 \pm 79}{22} \]
Calculating the two possible values:
1. \( y = \frac{66}{22} = 3 \)
2. \( y = \frac{-92}{22} \) (not valid)

Thus, \( y = 3 \).

### Step 10: Find \( x \)
Using \( y = 3 \) in the equation \( x = y + 2 \):
\[ x = 3 + 2 = 5 \]

### Step 11: Calculate the sum of the digits
Now, we find the sum of the digits:
\[ x + y = 5 + 3 = 8 \]

### Final Answer
The sum of the digits is:
\[ \boxed{8} \]