After how many decimal places will the decimal representation of the rational numbers `229/(2^2 times 5^7)` terminate?

Answer

To determine how many decimal places the decimal representation of the rational number \( \frac{229}{2^2 \times 5^7} \) will terminate, we can follow these steps:

### Step 1: Identify the factors of the denominator
The denominator is given as \( 2^2 \times 5^7 \).

### Step 2: Equalize the powers of 2 and 5
To make the denominator a power of 10 (which is \( 2^n \times 5^n \)), we need to equalize the powers of 2 and 5. The smaller power is \( 2^2 \), so we will multiply the numerator and denominator by \( 2^{(7-2)} = 2^5 \) to match the power of 5.

### Step 3: Modify the numerator and denominator
Now, we multiply the numerator and denominator by \( 2^5 \):
\[ \frac{229 \times 2^5}{2^2 \times 5^7 \times 2^5} = \frac{229 \times 32}{2^7 \times 5^7} \]
Calculating the numerator:
\[ 229 \times 32 = 7328 \]
So we have:
\[ \frac{7328}{2^7 \times 5^7} \]

### Step 4: Rewrite the denominator
The denominator can be rewritten as:
\[ 2^7 \times 5^7 = (2 \times 5)^7 = 10^7 \]
Thus, we can rewrite the fraction as:
\[ \frac{7328}{10^7} \]

### Step 5: Determine the decimal representation
The expression \( \frac{7328}{10^7} \) indicates that the decimal representation will be \( 7328 \) divided by \( 10^7 \), which means moving the decimal point 7 places to the left.

### Final Result
Therefore, the decimal representation of the rational number \( \frac{229}{2^2 \times 5^7} \) will terminate after **7 decimal places**.