The fraction `(7)/(q)` has a terminating decimal expansion. Which of these CANNOT be q?

Options

A. `8 xx 2`
B. `8 xx 3`
C. `8 xx 4`
D. `8 xx 5`

Answer

To determine which value of \( q \) cannot make the fraction \( \frac{7}{q} \) have a terminating decimal expansion, we need to understand the conditions under which a fraction has a terminating decimal expansion.

### Step-by-Step Solution:

1. **Understanding Terminating Decimals**:
A fraction \( \frac{a}{b} \) has a terminating decimal expansion if the denominator \( b \) (in its simplest form) can be expressed as \( 2^m \times 5^n \), where \( m \) and \( n \) are non-negative integers. This means that \( b \) can only have the prime factors 2 and 5.

2. **Analyzing \( q \)**:
We need to analyze the given options for \( q \) to see if they can be expressed in the form \( 2^m \times 5^n \).

3. **Checking Each Option**:
- **Option 1: \( 8 \times 2 \)**:
- \( 8 = 2^3 \), so \( 8 \times 2 = 2^4 \).
- This can be expressed as \( 2^m \) (with \( m = 4 \) and \( n = 0 \)).
- Thus, \( \frac{7}{8 \times 2} \) has a terminating decimal.

- **Option 2: \( 8 \times 3 \)**:
- \( 8 = 2^3 \) and \( 3 \) is a prime factor that is neither 2 nor 5.
- Thus, \( 8 \times 3 = 2^3 \times 3^1 \).
- Since it contains a factor of 3, it cannot be expressed as \( 2^m \times 5^n \).
- Therefore, \( \frac{7}{8 \times 3} \) does not have a terminating decimal.

- **Option 3: \( 16 \)**:
- \( 16 = 2^4 \).
- This can be expressed as \( 2^m \) (with \( m = 4 \) and \( n = 0 \)).
- Thus, \( \frac{7}{16} \) has a terminating decimal.

- **Option 4: \( 8 \times 5 \)**:
- \( 8 = 2^3 \) and \( 5 \) is a prime factor.
- Thus, \( 8 \times 5 = 2^3 \times 5^1 \).
- This can be expressed as \( 2^m \times 5^n \) (with \( m = 3 \) and \( n = 1 \)).
- Therefore, \( \frac{7}{8 \times 5} \) has a terminating decimal.

4. **Conclusion**:
The option that cannot be \( q \) such that \( \frac{7}{q} \) has a terminating decimal expansion is **Option 2: \( 8 \times 3 \)**.