Question
how many digits are repeating cycle of 17/26
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GPT-4o mini
To find the length of the repeating cycle in the decimal representation of the fraction \( \frac{17}{26} \), we first need to determine if the denominator can be simplified.
The denominator \( 26 \) can be factored into \( 2 \times 13 \). Since \( 2 \) is a prime factor of the denominator and it does not contribute to the repeating part, we focus on \( 13 \).
The repeating cycle length of a fraction \( \frac{a}{b} \) in simplest form is determined by the denominator's factors of \( 10 \) (where \( 10 = 2 \times 5 \)).
In this case, \( 13 \) is not a factor of \( 10 \), hence it will result in a repeating decimal.
Now, we find the order of \( 10 \) modulo \( 13 \), which is the smallest integer \( d \) such that \( 10^d \equiv 1 \mod 13 \). We can calculate the powers of \( 10 \) modulo \( 13 \):
- \( 10^1 \mod 13 = 10 \)
- \( 10^2 \mod 13 = 100 \mod 13 = 9 \)
- \( 10^3 \mod 13 = 1000 \mod 13 = 12 \)
- \( 10^4 \mod 13 = 10000 \mod 13 = 3 \)
- \( 10^5 \mod 13 = 100000 \mod 13 = 4 \)
- \( 10^6 \mod 13 = 1000000 \mod 13 = 1 \)
The order is \( 6 \) because \( 10^6 \equiv 1 \mod 13 \).
Thus, the repeating cycle of the fraction \( \frac{17}{26} \) has a length of **6** digits.
The denominator \( 26 \) can be factored into \( 2 \times 13 \). Since \( 2 \) is a prime factor of the denominator and it does not contribute to the repeating part, we focus on \( 13 \).
The repeating cycle length of a fraction \( \frac{a}{b} \) in simplest form is determined by the denominator's factors of \( 10 \) (where \( 10 = 2 \times 5 \)).
In this case, \( 13 \) is not a factor of \( 10 \), hence it will result in a repeating decimal.
Now, we find the order of \( 10 \) modulo \( 13 \), which is the smallest integer \( d \) such that \( 10^d \equiv 1 \mod 13 \). We can calculate the powers of \( 10 \) modulo \( 13 \):
- \( 10^1 \mod 13 = 10 \)
- \( 10^2 \mod 13 = 100 \mod 13 = 9 \)
- \( 10^3 \mod 13 = 1000 \mod 13 = 12 \)
- \( 10^4 \mod 13 = 10000 \mod 13 = 3 \)
- \( 10^5 \mod 13 = 100000 \mod 13 = 4 \)
- \( 10^6 \mod 13 = 1000000 \mod 13 = 1 \)
The order is \( 6 \) because \( 10^6 \equiv 1 \mod 13 \).
Thus, the repeating cycle of the fraction \( \frac{17}{26} \) has a length of **6** digits.
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