The product of two consecutive odd numbers is 399. Find the lower of them.

A. 17

B. 19

C. 21

D. 23

Options

A. B
B. C
C. A
D. D

Answer

To solve the problem of finding the lower of two consecutive odd numbers whose product is 399, we can follow these steps:

### Step 1: Define the Variables
Let the lower odd number be \( x \). Since we are looking for consecutive odd numbers, the next odd number would be \( x + 2 \).

### Step 2: Set Up the Equation
The product of these two consecutive odd numbers can be expressed as:
\[ x \cdot (x + 2) = 399 \]

### Step 3: Expand the Equation
Expanding the left side gives:
\[ x^2 + 2x = 399 \]

### Step 4: Rearrange the Equation
To form a standard quadratic equation, we rearrange it:
\[ x^2 + 2x - 399 = 0 \]

### Step 5: Factor the Quadratic Equation
Now we need to factor the quadratic equation. We look for two numbers that multiply to \(-399\) and add to \(2\). The factors of \(-399\) that satisfy this condition are \(21\) and \(-19\). Thus, we can factor the equation as:
\[ (x + 21)(x - 19) = 0 \]

### Step 6: Solve for \( x \)
Setting each factor to zero gives us:
1. \( x + 21 = 0 \) → \( x = -21 \) (not valid since we are looking for positive odd numbers)
2. \( x - 19 = 0 \) → \( x = 19 \)

### Step 7: Identify the Lower Odd Number
Since \( x = 19 \), the lower of the two consecutive odd numbers is:
\[ \text{Lower odd number} = 19 \]

### Conclusion
Thus, the lower of the two consecutive odd numbers whose product is 399 is **19**.