What can you say about the prime factorisation of the denominators of the rational number `0.bar(1347)`

Answer

To determine the prime factorization of the denominator of the rational number \(0.\overline{1347}\), we can follow these steps:

### Step 1: Define the repeating decimal
Let \( x = 0.\overline{1347} \).

### Step 2: Eliminate the repeating part
Since the decimal repeats every 4 digits (1347), we can multiply \( x \) by \( 10^4 = 10000 \) to shift the decimal point:
\[ 10000x = 1347.\overline{1347} \]

### Step 3: Set up an equation
Now we have two equations:
1. \( x = 0.\overline{1347} \) (Equation 1)
2. \( 10000x = 1347.\overline{1347} \) (Equation 2)

### Step 4: Subtract the two equations
Subtract Equation 1 from Equation 2:
\[ 10000x - x = 1347.\overline{1347} - 0.\overline{1347} \]
This simplifies to:
\[ 9999x = 1347 \]

### Step 5: Solve for \( x \)
Now, divide both sides by 9999:
\[ x = \frac{1347}{9999} \]

### Step 6: Factor the denominator
Next, we need to find the prime factorization of the denominator \( 9999 \). We can start by dividing by small prime numbers:
1. \( 9999 \div 3 = 3333 \)
2. \( 3333 \div 3 = 1111 \)
3. \( 1111 \div 11 = 101 \) (since 1111 is divisible by 11)
4. \( 101 \) is a prime number.

So, the prime factorization of \( 9999 \) is:
\[ 9999 = 3^2 \times 11 \times 101 \]

### Conclusion
The prime factorization of the denominator \( 9999 \) contains prime factors other than \( 2 \) and \( 5 \). This confirms that the denominator of the fraction representation of the non-terminating repeating decimal \( 0.\overline{1347} \) has prime factors other than \( 2 \) and \( 5 \).

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