The decimal expansion of `(23)/(2^(5)xx 5^(2))` will terminate after how many places of decimal ?

Options

A. 2
B. 4
C. 5
D. 1

Answer

To determine how many places the decimal expansion of \(\frac{23}{2^5 \times 5^2}\) will terminate, we can follow these steps:

### Step 1: Identify the prime factorization of the denominator
The denominator is given as \(2^5 \times 5^2\).

### Step 2: Equalize the powers of 2 and 5
For a decimal to terminate, the denominator (after simplification) must be of the form \(2^m \times 5^n\) where \(m\) and \(n\) are non-negative integers. In our case, we have:
- \(2^5\) (which is already in the required form)
- \(5^2\) (which is also in the required form)

### Step 3: Make the powers of 2 and 5 equal
To equalize the powers of 2 and 5, we can multiply the denominator by \(5^3\) (to make the power of 5 equal to 5). We also need to multiply the numerator by \(5^3\) to keep the equation balanced.

So, we have:
\[ \frac{23 \times 5^3}{2^5 \times 5^2 \times 5^3} = \frac{23 \times 125}{2^5 \times 5^5} \]

### Step 4: Calculate the new numerator and denominator
Calculating the new numerator:
\[ 23 \times 125 = 2875 \]
The new denominator becomes:
\[ 2^5 \times 5^5 = 10^5 = 100000 \]

### Step 5: Write the fraction
Now we can write the fraction as:
\[ \frac{2875}{100000} \]

### Step 6: Determine the decimal expansion
To find the decimal expansion, we can divide \(2875\) by \(100000\):
\[ 2875 \div 100000 = 0.02875 \]

### Step 7: Count the decimal places
The decimal \(0.02875\) has 5 decimal places.

### Conclusion
Thus, the decimal expansion of \(\frac{23}{2^5 \times 5^2}\) will terminate after **5 places of decimal**.

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