The average of three consecutive odd numbers is 52 more than `1/3`rd of the largest of these numbers. What is the smallest of these numbers ?

Options

A. 79
B. 75
C. 81
D. 77

Answer

To solve the problem, we need to find the smallest of three consecutive odd numbers given that their average is 52 more than one third of the largest number. Let's denote the smallest of these three consecutive odd numbers as \( A \). The three consecutive odd numbers can then be represented as:

1. \( A \) (smallest)
2. \( A + 2 \) (middle)
3. \( A + 4 \) (largest)

### Step 1: Calculate the average of the three numbers
The average of these three numbers is calculated as follows:

\[ \text{Average} = \frac{A + (A + 2) + (A + 4)}{3} \]

Simplifying this:

\[ \text{Average} = \frac{3A + 6}{3} = A + 2 \]

### Step 2: Calculate one third of the largest number
The largest of the three numbers is \( A + 4 \). One third of this number is:

\[ \frac{1}{3}(A + 4) \]

### Step 3: Set up the equation based on the problem statement
According to the problem, the average is 52 more than one third of the largest number. This can be expressed as:

\[ A + 2 = \frac{1}{3}(A + 4) + 52 \]

### Step 4: Clear the fraction
To eliminate the fraction, multiply the entire equation by 3:

\[ 3(A + 2) = A + 4 + 156 \]

This simplifies to:

\[ 3A + 6 = A + 160 \]

### Step 5: Solve for \( A \)
Now, rearranging the equation to isolate \( A \):

\[ 3A - A = 160 - 6 \]
\[ 2A = 154 \]
\[ A = 77 \]

### Conclusion
The smallest of the three consecutive odd numbers is \( 77 \).