Question

How many digits are in the repeating cycle of 17 over 27
?

Answer 1

To find how many digits are in the repeating cycle of the decimal representation of the fraction \(\frac{17}{27}\), we first need to perform the division of 17 by 27.

Dividing 17 by 27, we get:

\[
\frac{17}{27} = 0.\overline{629}
\]

Now, we observe that the decimal representation of \(\frac{17}{27}\) is \(0.629629629...\) which indicates that the digits "629" repeat indefinitely.

To determine the length of the repeating cycle, we can analyze the decimal expansion. The repeating part "629" consists of 3 digits. Therefore, the repeating cycle has a length of 3.

To verify our result, we can consider another approach using the properties of fractions. The length of the repeating part of the decimal representation of \(\frac{a}{b}\) (in simplest form) can also be found by determining the order of 10 modulo \(b\) (the denominator after removing factors of 2 and 5):

1. First, we reduce 27: it is in simplest form with respect to 17 (since they have no common factors).
2. Next, we calculate \(b=27\) and find \(10^k \equiv 1 \mod 27\). The smallest \(k\) for which this holds gives the length of the repeating cycle.

### Calculate the repeating cycle:
1. We find the powers of 10 modulo 27:
- \(10^1 \equiv 10 \mod 27\)
- \(10^2 \equiv 100 \equiv 19 \mod 27\)
- \(10^3 \equiv 10 \times 19 \equiv 190 \equiv 1 \mod 27\)

Since we find \(k = 3\) at \(10^3 \equiv 1 \) modulo 27, this means the repeating cycle has a length of 3.

Thus, the final answer is that the repeating cycle of \(\frac{17}{27}\) has:

\[
\boxed{3}
\] digits.