To determine how many digits are in the repeating cycle of the fraction \( \frac{17}{27} \), we first need to perform the division of 17 by 27.
When we divide 17 by 27, we find that it is less than 1, so we can express it as a decimal. Using long division or a calculator, we compute:
\[ 17 \div 27 = 0.629629629... \]
We can see that the decimal starts repeating. The part that repeats is "629", which is a sequence of 3 digits.
To confirm that the repeating cycle has 3 digits, we can also look at the denominator, which is 27. To find the length of the repeating cycle of a fraction \( \frac{a}{b} \) where \( b \) is in its simplest form, you can look at the prime factorization of the denominator (after removing any factors of 2 and 5).
The prime factorization of 27 is \( 3^3 \). Since 27 contains only the factor 3, we can use the formula for the length of the repeating cycle given by the order of 10 modulo 27.
To find this order, we calculate the smallest \( d \) such that \( 10^d \equiv 1 \mod 27 \).
Calculating powers of 10 mod 27:
- \( 10^1 = 10 \)
- \( 10^2 = 100 \equiv 19 \mod 27 \)
- \( 10^3 = 1000 \equiv 1 \mod 27 \)
Thus, \( d = 3 \) is the smallest number such that \( 10^d \equiv 1 \mod 27\).
Thus, the length of the repeating cycle of \( \frac{17}{27} \) is indeed 3.
Therefore, the answer is:
\[ \boxed{3} \]