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Rational Numbers in Different Forms Practice
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How many digits are in the repeating cycle of 1727
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GPT-4o mini
To determine how many digits are in the repeating cycle of \( \frac{1}{1727} \), we can perform the long division of 1 by 1727 or use the formula involving the decimal representation of fractions.
The length of the repeating cycle of a fraction \( \frac{1}{d} \) (where \( d \) is an integer) can be found using the properties of the number \( d \). Specifically, it involves finding the smallest integer \( k \) such that \( 10^k \equiv 1 \mod d \).
In this case, since \( 1727 \) is not a prime number, we need to take into account its prime factorization.
The prime factorization of \( 1727 \) is \( 7 \times 13 \times 19 \).
The length of the repeating decimal will be influenced by the factors of 1727 which are not factors of 10 (in \( 10 = 2 \times 5 \)). The length is determined by the order of 10 modulo each of the prime factors.
We should compute:
- The order of \( 10 \mod 7 \)
- The order of \( 10 \mod 13 \)
- The order of \( 10 \mod 19 \)
To compute the order of \( 10 \) modulo each of these primes, we need the smallest \( k \) such that \( 10^k \equiv 1 \mod p \).
1. **Order of 10 modulo 7**:
- \( 10^1 \mod 7 = 3 \)
- \( 10^2 \mod 7 = 2 \)
- \( 10^3 \mod 7 = 6 \)
- \( 10^4 \mod 7 = 4 \)
- \( 10^5 \mod 7 = 5 \)
- \( 10^6 \mod 7 = 1 \)
Hence, the order is 6.
2. **Order of 10 modulo 13**:
- \( 10^1 \mod 13 = 10 \)
- \( 10^2 \mod 13 = 9 \)
- \( 10^3 \mod 13 = 12 \)
- \( 10^4 \mod 13 = 3 \)
- \( 10^5 \mod 13 = 4 \)
- \( 10^6 \mod 13 = 1 \)
Hence, the order is 6.
3. **Order of 10 modulo 19**:
- \( 10^1 \mod 19 = 10 \)
- \( 10^2 \mod 19 = 5 \)
- \( 10^3 \mod 19 = 12 \)
- \( 10^4 \mod 19 = 6 \)
- \( 10^5 \mod 19 = 3 \)
- \( 10^6 \mod 19 = 11 \)
- \( 10^7 \mod 19 = 16 \)
- \( 10^8 \mod 19 = 8 \)
- \( 10^9 \mod 19 = 4 \)
- \( 10^{10} \mod 19 = 2 \)
- \( 10^{11} \mod 19 = 1 \)
Hence, the order is 18.
Now, the length of the repeating cycle of the decimal expansion of \( \frac{1}{1727} \) is the least common multiple (LCM) of these orders (6, 6, and 18).
The LCM(6, 6, 18) is 18.
Thus, the number of digits in the repeating cycle of \( \frac{1}{1727} \) is 18.
The length of the repeating cycle of a fraction \( \frac{1}{d} \) (where \( d \) is an integer) can be found using the properties of the number \( d \). Specifically, it involves finding the smallest integer \( k \) such that \( 10^k \equiv 1 \mod d \).
In this case, since \( 1727 \) is not a prime number, we need to take into account its prime factorization.
The prime factorization of \( 1727 \) is \( 7 \times 13 \times 19 \).
The length of the repeating decimal will be influenced by the factors of 1727 which are not factors of 10 (in \( 10 = 2 \times 5 \)). The length is determined by the order of 10 modulo each of the prime factors.
We should compute:
- The order of \( 10 \mod 7 \)
- The order of \( 10 \mod 13 \)
- The order of \( 10 \mod 19 \)
To compute the order of \( 10 \) modulo each of these primes, we need the smallest \( k \) such that \( 10^k \equiv 1 \mod p \).
1. **Order of 10 modulo 7**:
- \( 10^1 \mod 7 = 3 \)
- \( 10^2 \mod 7 = 2 \)
- \( 10^3 \mod 7 = 6 \)
- \( 10^4 \mod 7 = 4 \)
- \( 10^5 \mod 7 = 5 \)
- \( 10^6 \mod 7 = 1 \)
Hence, the order is 6.
2. **Order of 10 modulo 13**:
- \( 10^1 \mod 13 = 10 \)
- \( 10^2 \mod 13 = 9 \)
- \( 10^3 \mod 13 = 12 \)
- \( 10^4 \mod 13 = 3 \)
- \( 10^5 \mod 13 = 4 \)
- \( 10^6 \mod 13 = 1 \)
Hence, the order is 6.
3. **Order of 10 modulo 19**:
- \( 10^1 \mod 19 = 10 \)
- \( 10^2 \mod 19 = 5 \)
- \( 10^3 \mod 19 = 12 \)
- \( 10^4 \mod 19 = 6 \)
- \( 10^5 \mod 19 = 3 \)
- \( 10^6 \mod 19 = 11 \)
- \( 10^7 \mod 19 = 16 \)
- \( 10^8 \mod 19 = 8 \)
- \( 10^9 \mod 19 = 4 \)
- \( 10^{10} \mod 19 = 2 \)
- \( 10^{11} \mod 19 = 1 \)
Hence, the order is 18.
Now, the length of the repeating cycle of the decimal expansion of \( \frac{1}{1727} \) is the least common multiple (LCM) of these orders (6, 6, and 18).
The LCM(6, 6, 18) is 18.
Thus, the number of digits in the repeating cycle of \( \frac{1}{1727} \) is 18.