To determine the number of digits in the repeating cycle of \( \frac{1}{1727} \), we can use the properties of repeating decimals.
The length of the repeating cycle of \( \frac{1}{n} \) can be found by determining the smallest integer \( d \) such that \( 10^d \equiv 1 , (\text{mod} , n) \) where \( n = 1727 \).
First, we need to find the prime factorization of \( 1727 \):
- Since \( 1727 \) is odd and does not end in 0 or 5, it is not divisible by 2 or 5.
- Testing small primes, we find that \( 1727 \) can be factored as \( 7 \times 13 \times 19 \).
Next, we calculate the least common multiple of the lengths of the cycles of each prime factor separately:
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For the prime \( 7 \):
- The order of \( 10 , (\text{mod} , 7) \) is 6, since \( 10^1 \equiv 3 \), \( 10^2 \equiv 2 \), \( 10^3 \equiv 6 \), \( 10^4 \equiv 4 \), \( 10^5 \equiv 5 \), and \( 10^6 \equiv 1 \).
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For the prime \( 13 \):
- The order of \( 10 , (\text{mod} , 13) \) is 6, since \( 10^1 \equiv 10 \), \( 10^2 \equiv 9 \), \( 10^3 \equiv 12 \), \( 10^4 \equiv 3 \), \( 10^5 \equiv 4 \), and \( 10^6 \equiv 1 \).
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For the prime \( 19 \):
- The order of \( 10 , (\text{mod} , 19) \) is 18, since it takes 18 powers to return to 1.
To find the repeating cycle for \( 1727 \), we need to take the least common multiple (LCM) of the orders found:
- LCM(6, 6, 18) = 18.
Thus, the length of the repeating cycle of \( \frac{1}{1727} \) is 18 digits.