Set A consists only of fractions with numerator of 1 and a denominator d such that `1ltdlt8`, where d is an integer. If Set B consists of the reciprocals of the fractions with odd denominators in Set A, then what is the product of all the numbers that the elements of either Set A or Set B?

Options

A. `(1)/(96)`
B. `(1)/(48)`
C. `(1)/(24)`
D. `1`

Answer

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

### Step 1: Identify Set A
Set A consists of fractions with a numerator of 1 and denominators \( d \) such that \( 1 < d < 8 \) where \( d \) is an integer. The possible integer values for \( d \) are 2, 3, 4, 5, 6, and 7.

Thus, Set A can be written as:
\[ \text{Set A} = \left\{ \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7} \right\} \]

### Step 2: Identify Set B
Set B consists of the reciprocals of the fractions in Set A that have odd denominators. The odd denominators in Set A are 3, 5, and 7. Therefore, the reciprocals are:
\[ \text{Set B} = \left\{ 3, 5, 7 \right\} \]

### Step 3: Combine Sets A and B
Now, we need to consider all the elements from both sets:
- From Set A: \( \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7} \)
- From Set B: \( 3, 5, 7 \)

### Step 4: Calculate the Product of All Elements
The product \( P \) of all elements in Set A and Set B can be expressed as:
\[ P = \left( \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} \times \frac{1}{6} \times \frac{1}{7} \right) \times (3 \times 5 \times 7) \]

### Step 5: Simplify the Product
Calculating the product of the fractions in Set A:
\[ \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} \times \frac{1}{6} \times \frac{1}{7} = \frac{1}{2 \times 3 \times 4 \times 5 \times 6 \times 7} \]
Calculating the product of the integers in Set B:
\[ 3 \times 5 \times 7 = 105 \]

Now combining these:
\[ P = \frac{1}{2 \times 3 \times 4 \times 5 \times 6 \times 7} \times 105 \]

### Step 6: Calculate the Denominator
Calculating \( 2 \times 3 \times 4 \times 5 \times 6 \times 7 \):
\[ 2 \times 3 = 6 \]
\[ 6 \times 4 = 24 \]
\[ 24 \times 5 = 120 \]
\[ 120 \times 6 = 720 \]
\[ 720 \times 7 = 5040 \]

So, the denominator is 5040.

### Step 7: Final Calculation
Now we have:
\[ P = \frac{105}{5040} \]

To simplify:
\[ \frac{105}{5040} = \frac{1}{48} \]

### Final Answer
Thus, the product of all the numbers that are elements of either Set A or Set B is:
\[ \boxed{\frac{1}{48}} \]