How many digits are in the repeating cycle of 1727

?(1 point)

1 answer

To find the number of digits in the repeating cycle of the decimal expansion of \( \frac{1}{1727} \), we can first find the prime factorization of the denominator (1727) and then determine the order of 10 modulo 1727.

Check for prime factorization:
- \( 1727 \) can be checked for divisibility by prime numbers.
- It turns out that \( 1727 \) is actually equal to \( 7 \times 13^2 \).
Denominator properties:
- The decimal expansion of \( \frac{1}{1727} \) will be repeating because \( 1727 \) is not made up solely of the factors 2 and 5.
- The length of the repeating cycle can be determined using the formula for periodic decimal expansions, which is connected to the order of 10 modulo the coprime factor of 1727.
Finding the order of 10 modulo \( n \):
- To find the length of the repeating decimal, we need to find the order of 10 mod \( 1727 \). This requires \( n = 1727 \) to be broken into its prime factors.
- We use the formula to find the least common multiple (LCM) of orders for each prime power.
Separate calculations:
- Calculate the order of 10 modulo \( 7 \) (which is 10):
  - The multiples of \( 7 \) are \( 1, 2, 3, \ldots, 6 \), and we find:
    - \( 10^1 \mod 7 = 3 \)
    - \( 10^2 \mod 7 = 2 \)
    - \( 10^3 \mod 7 = 6 \)
    - \( 10^6 \mod 7 = 1 \)
    - Therefore, the order of \( 10 \mod 7 \) is \( 6 \).
- Calculate the order of 10 modulo \( 13^2 \) (which is \( 169 \)):
  - The totient function \( \phi(169) = 169(1 - \frac{1}{13}) = 169 \times \frac{12}{13} = 156 \). We need the divisors of \( 156 \).
  - Testing values, we find eventually that \( 10^{12} \mod 169 = 1 \). Thus, the order of \( 10 \mod 169 \) is \( 12 \) after checking smaller factors.
Combine orders:
- To find the length of the repeating cycle for \( 1727 \), take the least common multiple:
  - LCM of \( 6 \) and \( 12 \): \[ \text{LCM}(6, 12) = 12 \]

Thus, the number of digits in the repeating cycle of \( \frac{1}{1727} \) is \( \boxed{12} \).